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VISUAL OPTICS LABORATORY
BASIC OPTICAL PRINCIPLES
Prof.Dr.A.Necmeddin YAZICI
GAZİANTEP UNIVERSITY
OPTİCAL and ACOUSTICAL ENGINEERING DEPARTMENT
http://opac.gantep.edu.tr/index.php/tr/
1
BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
• Light is electromagnetic wave which has two components (E and M) and its
speed is equal to 3.0x108 m/s in air or vacuum.
• Light waves that are out of phase are called incoherent (tutarsız), while light
composed of waves exactly in phase is termed coherent (uyumlu).
• EM radiation is emitted in discrete packages of energy referred to as photons or
quanta.
• As seen in below figure, wave propagation can be represented as a sine wave
that varies in time and space.
• We classify electromagnetic wave according to its wavelength.
• Wavelength is defined as the distance between two corresponding points on two
consecutive waves.
FIGURE: The electromagnetic wave train is sinusoidal in form, and has repeating crests (maximum +
amplitude) and troughs (minimum - amplitude).
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BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
• Amplitude (A) is the maximum displacement of the vibrations of the wave.
• The amplitude is proportional to the inverse of the distance and thus the intensity is
proportional to the square of the inverse distance.
• The propagation of light in space may be represented by two different concepts:
wave motion and wavefronts.
• Electromagnetic energy travels as transverse waves, in which components of the
wave vibrate perpendicular to the direction of propagation.
• Wave motion is used to explain many aspects of physical optics, such as
interference and diffraction.
FIGURE : Electrical and magnetic fields propagating through space as an electromagnetic wave.
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BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
• The relationship between the frequency f (the number of vibrations per second of
an EM wave train); the wavelength λ (the distance from one crest to another); and
the velocity of light v, in meters per second, is given by
v = f   f = v/λ
• The energy of a photon is directly proportional to its frequency; so electromagnetic
radiation with a higher frequency also has a higher energy level.
• The amount of energy in a photon is given by
E = hf = hv/
• where E is the amount of energy per photon and h is Planck’s constant.
• As the wavelength decreases, the amount of energy per photon increases.
• For this reason, the absorption of short-wavelength radiation by body tissues is
typically more damaging than the absorption of longer-wavelength radiation.
• The case with light waves is a bit different.
• The light waves need no material medium to travel.
• They can propagate in vacuum.
• Light is a non-mechanical wave.
4
BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
Example: Is light a wave or a particle?
The answer is neither and both: light has a number of physical properties, some
associated with waves and others with particles. In some experiments light acts as a
wave and in others it acts as a particle.
Example:
The wavelength of a sodium (Na) source in a vacuum is 589 nm. What is the
frequency of the Na source in a vacuum?
Substituting the known values into the following equation yields
Example:
Light from Na source (f = 5.09x1014 Hz) enters a material where it travels at 2x108
m/s. What is the wavelength of the light in the material?
5
BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
Example:
Red light, with a wavelength of 600 nm in a vacuum, enters a block of glass
whereupon the wavelength reduces to 380 nm. What is the velocity and period of
the light in the block of glass?
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BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
EXAMPLE: A Half-Wave Antenna
A half-wave antenna works on the principle that the optimum length of the antenna
is one-half the wavelength of the radiation being received. What is the optimum
length of a car antenna when it receives a signal of frequency 94.0 MHz?
The wavelength of the signal is
Thus, to operate most efficiently, the antenna should have a length of (3.19 m)/2 =
1.60 m. For practical reasons, car antennas are usually one-quarter wavelength in
size.
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BASIC OPTICAL PRINCIPLES
Light and the Electromagnetic Spectrum
Example:
A satellite exploring Jupiter transmits data to the earth as a radio wave with a
frequency of 200 MHz. What is the wavelength of the electromagnetic wave, and
how long does it take the signal to travel 800 million kilometers from Jupiter to the
earth?
Solution: Radio waves are sinusoidal electromagnetic waves traveling with speed
c. Thus
The time needed to travel 800x106 km = 8.0x1011 m is
8
Light and the Electromagnetic Spectrum
• Electromagnetic wavelengths range in scale from that of an atomic nucleus (gamma
rays) to that of a small planet (radio waves).
• However, the optical radiation lies between X-rays and microwaves in the
electromagnetic spectrum and is subdivided into seven wavebands.
• They are
•
•
•
•
•
•
•
UV,
IR,
Microwave,
FM,
Radio Waves,
AM, and
Long radio waves.
Figure 1: Electromagnetic Spectrum – Image source: Wikipedia
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Light and the Electromagnetic Spectrum
• Just below the visible spectrum from 1 to 380 nm lies ultraviolet (UV), while just above
from 780 nm to 1 mm lies infrared (IR).
Figure 1: Electromagnetic Spectrum – Image source: Wikipedia
• Ultraviolet C (UV-C), 200–280 nm;
• Ultraviolet B (UV-B), 280–315 nm;
• Ultraviolet A (UV-A), 315–380 nm;
• Infrared A (IRA), 780–1400 nm;
• Infrared B (IRB), 1400–3000 nm;
• Infrared C (IRC), 3000–10000 nm.
• Visible radiation, 380–780 nm;
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Light and the Electromagnetic Spectrum
• Only the visible region of the electromagnetic spectrum that makes it through our
corneas and is absorbed by our retinas is perceived as color and light.
• The cornea and sclera of the eye absorb essentially all the incident optical radiation
at very short wavelengths in the ultraviolet (UV-B and UV-C) and long wavelengths
in the infrared (IR-B and IR-C).
• The incident UV-A is strongly absorbed by the crystalline lens of eye while
wavelengths in the range 400–1400 nm (visible light and near infrared) pass through
the ocular media to fall on the retina.
• The wavebands between 350 nm in the UV and 441 nm in the visible spectrum are
potentially the most dangerous for causing retinal damage under normal
environmental conditions.
• The transmission of radiation by the ocular media is summarized in below table.
TABLE: Transmission of Radiation by the Ocular Media
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Light and the Electromagnetic Spectrum
• It is desirable that intraocular (göziçi) lenses filter out these wavelengths and protect
the retina.
• Each of these elicits (sebep olmak) biological reactions.
• The visible wavelengths stimulate the retinal photoreceptors giving the sensation of
light while the near infrared may give rise to thermal effects.
• Because the refractive surfaces of the eye focus the incident infrared radiation on
the retina, it can cause retinal damage, e.g. eclipse burns (güneş tutulması yanığı).
• When radiation is absorbed by an ocular tissue, various effects are produced by the
transfer of radiant energy to the molecules and atoms of the absorbing tissue.
• The absorbed energy can effect the visual apparatus in the following ways:
• The Thermal Effect.
• The Photochemical Effect
• Photoluminescence (Fluorescence):
• The crystalline lens is especially capable of visible fluorescence, when
illuminated by ultraviolet light.
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BASIC OPTICAL PRINCIPLES
• A common acronym (kısaltma) used to remember the order of colors in the visible
spectrum is ROY G BIV (red, orange, yellow, green, blue, indigo, and violet).
• White light is composed of all the wavelengths in the visible spectrum. Individual
wavelengths within the visible spectrum, by themselves, create different color
sensations as shown in Table.
• The wavelengths of the visible spectrum lie between 380 and 760 nm, with red
light at the longer end of the spectrum and violet light at the shorter end.
• These are the spectral colors.
TABLE: The UV, visible, and IR spectrum
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Light Sources
• Unfortunatelly, there are no sharp cutoffs at the upper and lower limits, so it is
difficult to determine the range of wavelengths included in the visible spectrum
with any degree of certainty.
• It varies with the level of illumination, the clarity of the crystalline lens of the eye,
and other factors relative to the observer.
• The energy content of ultraviolet light is greater than that of infrared radiation
(fu>fi).
Euv>Einf
• Light is emitted by a luminous (or primary) source, which generates the
radiation.
• This radiation is often produced by heat, and such sources include the sun,
incandescent light bulbs, etc.
• Other objects are visible because they reflect light from luminous sources.
• These objects are called secondary sources.
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Wavefronts and Rays
• In geometrical optics, there is a definite relationship between waves, wavefronts,
and rays.
• Below figure shows a point source emitting waves in all directions.
• If a line is drawn connecting the same position (first peak, for example) on all the
waves, the result would be an arc with its center at the source.
• This arc represents one wavefront.
Figure: The relationship between waves, wavefronts, and rays.
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Wavefronts and Rays
• A ray of light is an imaginary line drawn along the direction of travel of the light
beam.
• For example, a beam of sunlight passing through a darkened room traces out the
path of a light ray.
• The rays, corresponding to the direction of wave motion, are straight lines
perpendicular to the wave fronts.
• When light rays travel in parallel paths, the wave fronts are planes perpendicular
to the rays.
• Light travels in a straight-line path in a homogeneous medium, until it encounters
a boundary between two different materials.
Figure: The relationship between waves, wavefronts, and rays.
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Wavefronts
• So, the light waves from a luminous point source
travel in every direction and form spherical wave
fronts.
• Think of them as circles with radii equal to the
distance from the point source (Fig-1).
• The curvature of these wavefronts decreases as
the distance from the source increases.
• An arc with a longer radius is flatter than one with
a shorter radius (Fig-2).
• At infinity (where the radius of the arc is infinity),
the wavefronts are flat.
Figure-1: A point source of light emits
concentric waves of light. Light rays,
represented by arrows, are perpendicular
to the wavefronts.
Figure-2: The curvature of wavefronts becomes less as the
distance from the point source increases. They are arcs of a circle
whose center is the point source. At infinity, the wavefronts are flat.
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Wavefronts
• All rays emanating from a point source or a single point on an extended source
may be grouped to form a pencil of rays.
• Depending upon the direction the rays travel within the pencil, pencils may be
defined as convergent, divergent, or parallel.
Figure: On-axis and off-axis divergent and convergent pencils. Note that the divergent rays are traveling
away from each other and the radius of the wavefront is increasing. The convergent rays are aimed toward
an image point, and the wavefront radius is decreasing.
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Wavefronts
• In below figure; an infinite point source illuminates a pupil (also called an aperture
or stop) with parallel wavefronts (and rays).
• If the aperture is large, the majority of the rays pass through, and only the size of
the pencil is reduced.
• If, however, the aperture is sufficiently small to allow only one ray to pass through,
a single wavelet will pass through the aperture, and the incident parallel wavefront
will emerge as a spherical wavefront with rays diverging from the stop.
• This is diffraction.
Figure: Rectilinear propagation of light occurs with large pupils (left) where incident parallel rays emerge
parallel. With small pupils (right), diffraction occurs, and incident parallel light emerges as a spherical wave.
19
CURVATURE
• The curvature of a surface is inversely proportional to its radius of curvature, and
will increase in magnitude as the radius decreases in magnitude.
• This is demonstrated in below figure with two different circles.
• The circle with the smallest radius has more curvature, and vice versa.
• The curvature R is given simply by,
R=1/r
• where r is the radius of circle or radius of curvature of the surface.
• The unit of measurement for curvature is the reciprocal meter (m-1).
Example
A spherical surface has a 0.5-m radius of
curvature. What is the curvature of the
surface?
FIGURE: The angle θ represents the angle
turned over a unit length of arc. The circle with
the 0.5-m radius has twice as much curvature
as the circle with the 1.0-m radius.
R = 1/0.5  R = 2 m-1
∴ Curvature is 2 m-1.
20
VERGENCE
• We use the term vergence to describe the amount of
curvature of a given wave front.
• For point sources, curvature is greatest near the
source and diminishes with distance from the source.
• The more curved a wavefront is, the greater its
vergence.
• Likewise, the less curved it is, the less its vergence.
• Unlike regular curvature, however, the vergence of a
wave front is measured in units called diopters—
instead of reciprocal meters.
• In air, the vergence L of the wave front—in diopters
(abbreviated ‘D’)—is equal to the reciprocal of the
radius of curvature l of the wave front.
• It is given by the simple relationship
L=1/l
where l is the distance from the source measured in
meters.
FIGURE: Point A represents a
source with wave trains of light
propagating outward in every
direction. The wave front is the
spherical shell that envelops all of
these wave trains at a particular
distance from the source. Line AB
represents a ray originating from
point A and perpendicular to the
wave front at point B.
• The vergence in air is equal to the reciprocal of the radius of the wave front.
• Note: A point source will produce 1 diopter of vergence at a distance of 1 meter.
21
Vergence
• The sign (±) of the value of l identifies the type of vergence.
• This is because vergence can be either positive or negative:
• Positive (+) values for vergence will produce convergent wave fronts that
come to a point.
• Negative (-) values for vergence will produce divergent wave fronts that
spread apart (as if from a point).
• Zero (0) values for vergence will produce parallel wave fronts, with no
vergence
Figure: An object located at infinity
produces a parallel light rays.
FIGURE: Optical sign convention.
22
Vergence
• In below figure, the vergence of the light rays is away from a point source of light;
this is called negative vergence, or divergence.
Figure: Diverging light rays have negative vergence. At distances of 0.50, 1.00, and 2.00 m, the
vergence is −2.00, −1.00, and −0.50 D, respectively. The magnitude of the vergence (ignoring the sign)
decreases as the distance to the source increases.
23
Vergence
• When light rays come together to form an image; this is called positive vergence, or
convergence.
• Consider below figure, which shows light converging to a point focus.
• The vergence measured at distances of 10.00, 20.00, and 50.00 cm from this focus
point is +10.00, +5.00, and +2.00 D, respectively.
• Note that the vergence is greatest close to the focus point and decreases as the
distance increases.
Figure : Converging light rays have positive vergence.
24
Vergence
FIGURE: Vergence values for wavefronts diverging from a point source.
FIGURE: Vergence values for converging wavefronts.
25
Vergence
Example
Light is diverging (negative) from an
object located 2 m away. What is the
vergence of the wave front?
L =1/(-2)  L = -0.50 D
∴ Vergence is -0.50 D (divergence)
Example
A wave front is converging (positive)
upon an image point located 1/3 m away.
What is the vergence of the wave front?
L = 1/(1/3)  L = 3.00 D
∴ Vergence is 3.00 D (convergence)
Example: What is the vergence of a light source located infinitely far away?
Wave fronts become progressively flatter as the distance from their center of
curvature increases. Therefore, as the radius of the wave front becomes infinitely
long, the vergence approaches zero. Think of it this way: since the light rays are
neither diverging nor converging, the vergence is zero. The vergence L approaches
0 as the distance l approaches optical infinity (∞). For clinical purposes, we normally
consider distances greater than 20 ft (or 6 m) as infinitely far away. The actual
vergence past 6 m will be less than 0.17 D.
26
Vergence
EXAMPLE:
Consider diverging light traveling to the right as shown in below figure. The
vergence of the light at point A is -10.00 D. What is the vergence of the light at
point B, which is 15 cm downstream from A?
FIGURE: Downstream vergence example.
The magnitude of the radius of curvature of the wavefront at A is

From above figure, the radius of curvature of the wavefront at B is greater in
magnitude than that at A, or

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Vergence
Example (Cont.):
The magnitude of the vergence VB of the wavefront at B is then
Since the light is diverging, VB is minus, or
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Vergence
EXAMPLE:
In shown in below figure, the vergence of light at position C is -2.00 D. What is
the vergence at position D, which is 30 cm upstream from position C?
FIGURE: Upstream vergence example.
The magnitude of the radius of curvature of the wavefront at position C is
From above figure, the radius of the wavefront at D is shorter than the radius of
the wavefront at C, or

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Vergence
Example (Cont.):
Then
Since the light is diverging, the vergence is negative and
QUESTION:
The vergence of light at point C is +18.12 D. What is the vergence of the light at
point B, which is 36.75 cm upstream? Try doing a quick sketch to convince
yourself that VB should be positive and less than +18.12 D. Then verify that the
vergence at position B is +2.37 D.
30
Vergence
EXAMPLE:
Consider converging light traveling to the right (see below figure). The vergence of
the light at point A is +8.00 D. What is the vergence at point B, which is 5 cm
downstream from A?
FIGURE: Downstream vergence example.
The magnitude of the radius of curvature of the wavefront at A is
From the sketch, the magnitude of the radius of curvature at B is
31
Vergence
Example (Cont.):
Then
Since the light is converging at point B, the vergence is positive and
EXAMPLE:
As a final example in this section, consider light at position A with a vergence of
+25.00 D (see below figure). What is the vergence of the light at a position 15 cm
downstream?
FIGURE: Downstream vergence example for point image between wavefronts.
32
Vergence
Example (Cont.):
The magnitude of the radius of curvature of the wavefront at A is
From the sketch, the magnitude of the radius of curvature of the diverging
wavefront at B is
The magnitude of the vergence at B is
Since the wavefront is diverging
33
Vergence
• Vergence is an important concept used in deriving image-object relationships.
• Vergence is related to the radius of the wavefront.
• If the radius is increasing as the wave propagates, the wavefront is divergent and
is denoted as having a negative vergence.
• If the radius of the wavefront is decreasing as the wave propagates, the wavefront
is converging and has a positive vergence.
• In previous slayts, the vergence is defined as the reciprocal of the radius of the
wavefront.
• This is true in most cases; however, if the wave is traveling in a medium other
than air, the decrease in velocity (and the change in vergence) must be taken into
account.
• When the object is located in a primary medium other than air, the medium’s
refractive index increases the absolute value of the vergence.
34
Vergence
• This is called the reduced vergence, and is defined as follows:
Example:
What is the vergence of the light in the water at a distance of 28 cm from the point
source?
According to the sign convention,
R = -28 cm
and then


Note that the vergence is negative as it should be for diverging light.
35
Vergence
• The term reduced distance, either in object space or in image space, is the actual
distance divided by the index of refraction of the medium, such as
l/n or l’/n’
• Reduced vergence is the reciprocal of reduced distance. That is,
1/(l/n) or 1/(l’/n’)
• For an object located in a medium n, the absolute value of the object (incident)
vergence is given by the following relationship:
• The image (emergent) vergence is calculated as
• L and L' are examples of reduced vergence, expressed in terms of the diopter, (l)
is defined as the distance from the refracting surface to the object, (l' ) is defined
as the distance from the refracting surface to the image.
36
Vergence
• When measuring left to right, the distance is positive; measuring right to left, the
distance is negative.
• Arrows should be used to indicate the direction of measurement.
• As shown in below figure, the space in front of the surface (to the left of) may be
called object space and the space behind the surface (to the right of) may be
called image space.
• To work problems properly, sign conventions must be followed.
FİGURE: Space, image, and object definitions.
37
REAL AND VIRTUAL OBJECTS AND IMAGES
• Objects and images are real if the rays coming from the object or going to the
image actually intersect.
• If the rays intersect when extended backwards or forward the image and object,
so formed, are virtual.
Figure: A real image is formed when two rays actually intersect.
Figure: A virtual image is produced when two rays from source S must
be projected back in order to intersect at S'.
38
REAL AND VIRTUAL OBJECTS AND IMAGES
• Although, it is essential to adopt a sign convention for optical calculations, no
one sign convention is used universally.
• The sign convention used here is:
• 1. Light from the object initially travels from left to right.
• 2. Distances measured in the direction that light initially travels (from left to
right) are positive.
• 3. Distances measured opposite the direction that light initially travels are
negative.
39
REAL AND VIRTUAL OBJECTS AND IMAGES
• The object distance u is the step from A to M, that is, u = AM; the image distance
u' = AM' is the step from A to M‘.
• In below figure-a the object is real and the image is virtual.
• Both the object and image distances are negative.
• In figure-b the object is virtual, the image is real, and both object and image
distances are positive.
Figure: Paraxial object and image distances for refraction at a plane surface, (n'>n). a) Real object, virtual
image, b) Virtual object, real image.
40
Vergence
Example: What is the vergence for the object in below figure that is located in water
rather than air?

 Since the rays are diverging, we have L = −4.03 D
41
Vergence
Example:
What is the vergence of the rays forming the virtual image in below figure?

 L’ = -3.80 D
Since the rays that form a virtual image are diverging, their vergence must be
designated with a minus sign as follows: L′ = −3.80 D
42
Vergence
• In vergence relationship, the vergence of the object rays is added to the dioptric
power of the refractive surface to give us the vergence of the image rays.
• This can be expressed as the Gaussian Imaging Equation:
Object Vergence + Surface Power = Image Vergence
• Designating the object vergence as L, the surface power as F, and the image
vergence as L′, we have
L + F = L′
or
L′ = L + F
where:
L' = the emerging reduced vergence leaving the surface after refraction
L = the incident reduced vergence stricking the surface
F = the power of the surface
• From this equation, the relationship between the object position, image position,
and power of the surfaces may be calculated.
• The type of object or image (real or virtual) and the resulting type of vergence
(parallel, convergent, divergent) must be known before the equation can be
used properly.
• You must be careful to follow the established sign convention.
43
Vergence
• The vergence (paraxial) relationship is convenient for locating images and
determining magnification.
• The lateral magnification (yatay büyütme) is given by
• When a thin lens is located in air (or any other one medium), lateral magnification
can also be found with the following equation:
• It bears repeating that this relationship is valid only when the medium is the same on
both sides of a thin lens.
• The total lateral magnification of the thick lens is calculated by multiplying the
lateral magnification produced by the first surface by the lateral magnification
produced by the second surface:
Total lateral magnification = (ML for first surface) x (ML for second surface)
44
REFRACTION of LIGHT
• Light waves travel through transparent media at different speeds.
• The speed of light in a transparent material—such as air, water, or glass—differs
from the speed of light in vacuum.
• We can recognize this immediately by recalling the theoretical formula for the
speed of light derived from Maxwell’s equations,
• We know that in a material with given dielectric characteristics, the quantity 0 in
Maxwell’s equations must be replaced by 0, where  is the dielectric constant of
the material. Hence
• This is usually written as
• so
• The quantity n is called the index of refraction of the material.
• The index of refraction of any material is larger than 1; the speed of light in the
material is less than the speed of light in vacuum.
45
Index of Refraction
• The index of refraction (n) is a measure of how much the speed of light is altered as
it enters a given media relative to the speed of light in air.
• The index of refraction (n) can be calculated using the following equation:
n = speed of light in vacuum or (in air) / speed of light in selected material
• Except for air (and vacuums) which has a refractive
index of 1, the refractive index of most substances is
greater than unity (n > 1).
• The absolute refractive index of any material can be
determined using a refractometer.
• With the wave speed v = c/n, the relation between
frequency and wavelength becomes


• Thus the wavelength of an electromagnetic wave is
shorter in a material than in empty space.
• For example, if a wave penetrates from vacuum or
from air into water, its speed is reduced by a factor of
1.33, but its frequency remains constant.
Figure: As the wave moves
from medium 1 to medium 2, its
wavelength changes, but its
frequency remains constant.
46
REFRACTION of LIGHT
• A light wave travels through three transparent materials of equal thickness.
• Rank in order, from largest to smallest, the indices of refraction na, nb, and nc.
47
REFRACTION of LIGHT
• For example, blue light with a wavelength of 350 nm travels in a medium at a
slower velocity than red light with a wavelength of 750 nm.
• As light moves from one transparent medium to another, at any angle other than
perpendicular to the material surface, the change in speed will also result in a
change in direction.
Figure: The straw (kamış) seems to be broken, due to refraction of light as it emerges into the air.
48
REFRACTION of LIGHT
• The change in the direction of light is called refraction.
• The greater the change in speed, the greater the magnitude of refraction becomes.
• When going from a lower index medium to a higher index medium that is more
dense, such as from air to a piece of glass, the velocity is reduced.
• Refraction is principle that allows the creation of optical lenses that alter the path or
focus of light.
Question:
If beam 1 is the incoming beam in active below figure, which of the other four
beams are due to reflection? Which are due to refraction?
49
Refraction of Light
• Refraction does not always occur when light travels from one optical medium to
another.
• Although there is a change in the index of refraction as the light rays travel from the
primary medium (air) to the optically denser secondary medium (glass), the angle of
incidence is zero and refraction does not occur.
• As illustrated in Figure-B, the same holds true when light rays are directed toward
the center of curvature of the spherical surface of a glass rod; rays strike
perpendicular to the surface and are not refracted.
Figure A. Parallel light rays that strike a plane (i.e., flat) glass surface perpendicular to its surface are not
deviated. B. Similarly, rays headed toward the center of curvature (C) of a spherical glass surface strike the
surface perpendicular to its surface and are not deviated. A spherical surface is a section of a sphere. Its
radius, which is frequently referred to as the radius of curvature, is the distance from the surface to the center
of curvature.
50
Refraction of Light
• On the other hand, when the rays are going from a lower-index medium to a
higher-index medium that is more dense, such as from air to a piece of glass, the
rays of light are shifted toward the normal.
• When going from a higher index medium to a lower index medium that is less
dense, such as from a piece of glass to air, the reverse occurs and the rays of light
are shifted away from the normal of the surface.
FIGURE: The wave fronts of wave train 1 enter the glass medium perpendicularly, and are slowed down
uniformly. There is no change in direction. The wave fronts of wave train 2 enter the glass obliquely
(yandan): Side A of the approaching wave fronts strikes the glass before side B, causing side A to slow first.
As a result, the wave train is refracted, or bent, as it enters the medium.
51
Refraction of Light
Figure: Refraction at a low-high and high-low index interface.
52
Index of Refraction
• Below table gives the indices of refraction for some
representative substances. The values are listed for a
particular wavelength of light, because they vary slightly with
wavelength. (This can have important effects, such as colors
produced by a prism.)
• Note that for gases, n is close to 1.0.
• This seems reasonable, since atoms in gases are widely
separated and light travels at c in the vacuum between
atoms.
• It is common to take n = 1 for gases unless great precision is
needed.
• Although the speed of light v in a medium varies considerably
from its value c in a vacuum, it is still a large speed.
53
Index of Refraction
• The following are examples of n in various media:
Air: 1.00
Water: 1.33
Aqueous Humor: 1.33
Vitreous Humor: 1.33
Cornea: 1.37
Crystalline Lens: 1.42
CR-39 Plastic: 1.49
Crown Glass: 1.52
Polycarbonate: 1.58
Trivex: 1.53
High Index Plastics: 1.60 to 1.74
High Index Glass: 1.60 to 1.80
54
Index of Refraction
Example: Speed of Light in Matter
A ray of light travels at a velocity of 200,000 km/s through a particular lens material.
What is the index of refraction for that lens material?
Solution: n=300.000/200.000  n = 1.500
∴ Index of refraction is 1.500
Example: Speed of Light in Matter
Calculate the speed of light in zircon, a material used in jewelry to imitate diamond.
Solution: The speed of light in a material, v , can be calculated from the index of
refraction n of the material using the equation n = c/v. The index of refraction for
zircon is given as 1.923 and c is given in the equation for speed of light. Entering
these values in the expression gives

Example:
The wavelength of light coming from a sodium source is 589 nm. What will be its
wavelength in water ? Refractive index of water = 1.33.
Solution : The wavelength in water is  = 0 /n, where 0 is the wavelength in
vacuum and n is the refractive index of water. Thus,
 = 589/1.33 = 443 nm
55
Index of Refraction
EXAMPLE:
A light wave of wavelength 550 nm in vacuum enters a plate of glass of index of
refraction n 1.52.What is the speed of the light in the glass? What is the wavelength
of the light in the glass? What is the frequency of the wave in the glass?
SOLUTION: The speed of light in the glass is
The wavelength in the glass is
The frequency in the glass is the same as the frequency in vacuum,
Alternatively, we can calculate this frequency from the speed and the wavelength in
the glass, with the same result:
56
Index of Refraction
Example:
In an unknown material, light with a wavelength of 589 nm travels at 2x108 m/s.
a-) What is the index of refraction of the material?
b-) What is the wavelength of light in the material?
Solution: a-)
The index of refraction has no units because the m/s units cancel in the ratio.
b-)
In the unknown material, the frequency remains constant and the velocity changes.
Using the new velocity and the calculated frequency, the new wavelength is
determined by solving as;
57
Index of Refraction
Example: Light traveling through glass
Orange light with a wavelength of 600 nm is incident upon a 1.00-mm-thick glass
microscope slide.
a. What is the light speed in the glass?
b. How many wavelengths of the light are inside the slide?
Solution: a. The index of refraction of glass is nglass = 1.50. Thus the speed of light
in glass is
b. The wavelength inside the glass is
N wavelengths span a distance d = N , so the number of wavelengths in d = 1.00
mm is
The fact that 2500 wavelengths fit within 1 mm shows how small the wavelengths of
light are.
58
Index of Refraction
Example:
The wavelength of yellow light in vacuum is 600 nm. (a)What is the speed of this
light in vacuum and water? (b) What is the frequency of this light in vacuum and
water? (c) What is the wavelength of this light in water?
Solution: (a) The speed of the yellow light in vacuum (n=1) and water (n=1.333)
can be obtained by using the following equation as follows:

(b)We use the equation v = λ f to prove that the frequency of the yellow light in
vacuum and water is the same, as follows:
(c) we can calculate the wavelength of the yellow light in water as follows:
59
Fermat's Principle
• Keep in mind the definition of the index of refraction (n = c/v), the time (t) for light
to travel a distance (d) in a medium with an index (n) may be expressed by:
• The term nd is referred to as the optical distance or optical path length.
• From above equation; it is seen that the time to travel through some medium is
the optical path length divided by the speed of light in a vacuum.
• To calculate the time to travel through several media, sum the time to travel in
each medium:
• Principle of Least Time is the path traveled by light from one point to another in
a medium will be the path that requires the least amount of time.
• This is also called Fermat's Principle.
60
Fermat's Principle
Example:
What is the velocity of Na light in glass with an index of 1.50? How far will light travel
in air in 10-8 seconds? Compare this distance to the distance light will travel in glass
for the same time interval. What is the optical path length in each case?
The distance light travels in 10-8 seconds is calculated by multiplying the velocity
times the time:
Thus the light in this glass slows to 2/3 the velocity and travels 2/3 the distance
when compared to light in air. The optical path length for these distances is
Thus for the same time interval, light travels the same optical path length.
61
Fermat's Principle
Example:
As shown in below figure, light travels normal through several plane interfaces of
different indices: 3cm in index 1.33, 5cm in index 1.55, 10cm in index 1.70. How
much time is required to travel from the first through the last interface?
62
Dispersion
• Since the velocity of light varies as a function of wavelength in media other than a
vacuum, the index of refraction also varies as a function of wavelength in these
media.
• This relationship is dependent upon the components that make up the optical
material.
• A plot of index versus wavelength is shown in below figure for a representative
material.
• This plot is referred to as a dispersion curve.
• From this fgure, the index for
shorter wavelengths is greater
than for longer wavelengths, i.e.,
the
index
of
blue
light
(approximately 1.60 at 400 nm)
is greater than the index for red
light (approximately 1.47 at 700
nm).
Figure: A dispersion curve showing the variation in
index as a function of wavelength.
63
Dispersion
Table: Index of Refraction n in Selected Media at Various Wavelengths
Figure: A schematic representation of the dispersion of white light. The violet color is bent more than the
red color. (a) Dispersion in a glass block. (b) Dispersion in a prism
64
Dispersion
• In defining the index of refraction of optical media, one wavelength (589 nm or the
dominant wavelength of a sodium source) has been used as the standard.
Table: Wavelengths and sources of some prominent Fraunhofer lines.
65
Dispersion
• In this figure, the index of refraction at 589 nm is 1.50. In this and most other
optics texts, the index of refraction is defined for the standard wavelength of 589
nm unless otherwise stated.
• The indices listed in this figure are for this standard wavelength.
• A change in the standard wavelength to He D-line (587 nm) produced by a laser
has been proposed recently.
Figure: A dispersion curve showing the variation in index as a function of wavelength.
66
Dispersion
• The index of refraction of an optical glass depends on the wavelength of light.
• The index corresponding to a particular wavelength is identified with a subscript.
• Thus, nd (nD), nF and nc correspond to indices in yellow, blue and red light.
• For example, a common optical glass called BK7 has indices nd= 1.5168, nF =
1.52238 and nc =1.51432.
• When no subscript is used, the index in yellow helium light is understood.
• Thus, the index of BK7 is written simply as n = 1.5168.
• Indices of refraction of optical glasses vary from 1.44 to 2.04.
Table: Index of Refraction n in Selected Media at Various Wavelengths.
67
Mirages
• When the index of refraction of a medium changes gradually, the refraction is
continuous, leading to a gradual bending of the light.
• An interesting example of this is the formation of a mirage.
• On a hot and sunny day, the surface of exposed rocks, pavement, and sand often
gets very hot.
• In this case there is often a layer of air near the ground that is warmer, and
therefore less dense, than the air just above it.
• The speed of any light wave is slightly greater in this less dense layer, so a light
beam passing from the cooler layer into the warmer layer is bent.
• Figure-a shows the light from a tree when all the surrounding air is at the same
temperature.
• The wavefronts are spherical, and the rays are straight lines.
Figure: A mirage. (a) When the air is at a uniform temperature, the wave fronts of the light from the tree are
spherical.
68
Mirages
• In Figure-b, the air near the ground is warmer, resulting in the wavefronts traveling
faster there.
• The portions of the wavefronts near the hot ground get ahead of the higher
portions, creating a nonspherical wavefront and causing a curving of the rays.
• Thus, the two rays shown initially heading for the ground are bent upward.
Figure: A mirage. (b) When the air near the ground is warmer, the wavefronts are not spherical and the light
from the tree is continuously refracted into a curved path.
69
Mirages
• As a result, the viewer sees an image of the tree looking as if it were reflected off a
water surface on the ground.
• When driving on a hot sunny day, you may have noticed apparent wet spots on the
highway ahead that disappear as you approach them.
• These mirages are due to the refraction of light from the sky by a layer of air that
has been heated due to its proximity to the hot pavement.
Figure: A mirage. (c) Apparent reflections of motorcycles on a hot road.
70
Dispersion
Example:
Using the dispersion curve in below figure, determine the velocities for 400 nm, 589
nm, and 700 nm. How does the wavelength relate to the velocity and index?
These calculations show that as the
wavelength increases, the velocity
increases and the index of refraction
decreases.
71
Snell’s Law
• Snell’s law of refraction is fundamental to the
study of optics.
n ⋅ sin(i) = n′ ⋅ sin(i′)
• Snell’s law is illustrated in below figure for another
parallel block of glass.
• From the formula, we can also conclude that:
• If n > n' then i < i‘
• If n < n' then i > i'
Example:
A ray of light strikes a lens material with a 1.523 refractive index, at a 30° angle of
incidence in air (which has a refractive index of 1). What is the angle of refraction?
n ⋅ sin(i) = n′ ⋅ sin(i′)  1⋅ sin 30°= 1.523⋅ sini′  sini′ = 0.3283  i ′ = 19.17°
∴ Angle of refraction is 19.17º.
Question: What happens if the angle of incidence exceeds the critical angle?
72
Snell’s Law
Example: A ray strikes an air-water interface at an incident angle of 30°. What is the
angle of refraction? For the same incident angle, what would be the refracted angle
at a water-air interface?
Use Snell's Law to calculate the refracted angle. For the air-water interface


For the water-air interface


Example:
A refracted ray leaves normal to a crown glass-air interface. What is the angle of
incidence?
Apply Snell's Law. Note that the sin 0° = 0°


The incident angle is 0° and normal to the surface. A normal incident ray yields a
normal refracted ray, and the ray does not change direction. Since the velocity
changes, there is refraction.
73
Snell’s Law
EXAMPLE:
What will be the displacement of a ray through a 100-mm thick plate of index 1.5,
which is tilted 40°?


Figure:Path of a ray through a parallel plate in a
uniform medium and Displacement of the ray.
74
Snell’s Law
Example:
(a) In below figure, a beam of monochromatic light
reflects and refracts at point A on the interface between
material 1 with index of refraction n1 = 1.33 and material
2 with index of refraction n2 = 1.77. The incident beam
makes an angle of 50° with the interface. What is the
angle of reflection at point A? What is the angle of
refraction there?
The angle of incidence θ1 is not the given 50° but is
θ1 = 90° - 50° = 40°.
Thus, the angle of reflection at A is
θ1‘ = θ1 = 40.

Figure: (a) Light reflects and refracts at point A on the interface between
materials 1 and 2. (b) The light that passes through material 2 reflects and
refracts at point B on the interface between materials 2 and 3 (air). Each
dashed line is a normal. Each dotted line gives the incident direction of
travel.
75
Snell’s Law
Example (Cont.):
(b) The light that enters material 2 at point A then
reaches point B on the interface between material 2 and
material 3, which is air, as shown in figure-b.The
interface through B is parallel to that through A. At B,
some of the light reflects and the rest enters the air. What
is the angle of reflection? What is the angle of refraction
into the air?
The angle of reflection at B is
θ2‘ = θ2 = 28.88 ≈ 29.

Figure: (a) Light reflects and refracts at point A on the interface between
materials 1 and 2. (b) The light that passes through material 2 reflects and
refracts at point B on the interface between materials 2 and 3 (air). Each
dashed line is a normal. Each dotted line gives the incident direction of
travel.
76
Snell’s Law
Example:
A light ray of wavelength 589 nm traveling through air is incident on a smooth, flat
slab of crown glass at an angle of 30.0° to the normal.
(a) Find the angle of refraction.
(a) Find the speed of this light once it enters the glass.

(b) What is the wavelength of this light in the glass?
77
Snell’s Law
Example:
A light beam passes from medium 1 to medium 2, with the latter medium being a
thick slab of material whose index of refraction is n2 (see figure). Show that the
beam emerging into medium 1 from the other side is parallel to the incident beam.
What if the thickness t of the slab is doubled? Does the offset distance d also
double?
Therefore, θ3 = θ1 and the slab does not alter
the direction of the beam.
78
Snell’s Law
Example:
When sunlight passes through atmospheric ice crystals (see figure) that have their
hexagonal cross sections horizontal, it turns a ray from its original direction of travel
by an angle that depends on the crystal's orientation. The minimum turning angle
θdog is the angle to the left or right of the Sun at which you can see a sun dog. The
index of refraction of ice is 1 .3I.Find θdog.
79
Critical Angle
• For a high-low index interface (n > n'), the refracted ray bends away from the
normal, and therefore the refracted angle is greater than the incident angle.
• As the incident angle increases, the refracted angle also increases but always
remains larger than the incident angle (see below figure).
• Here there is a physical limit of 90o for the refracted angle (i.e., the refracted ray
leaves along the surface of the interface).
• The incident angle that yields a refracted angle of 90° is called the critical angle
and is labeled θC.
Figure: For high-low index interfaces, the refracted angle is greater than the incident angle. The limiting
incident ray has an incident angle (called the critical angle) that yields a 90° refracted angle. Rays with
incident angles greater that the critical angle are internally reflected.
80
Critical Angle
• For incident angles larger than the critical angle, the refracted angle is undefined.
• These incident angles must leave at angles greater than 90°, and therefore these
rays are reflected back into the same medium as the incident ray.
• Snell's Law may be used to determine the angle of reflection.
• For incident angles greater than the critical angle, rays undergo total internal
reflection.
• Internal reflection can also be used to calculate the index of refraction of an
unknown material.
• The critical angle may be calculated by using Snell's Law and solving for the
incident angle (θ) that yields a refracted angle (θ') of 90°·


81
Critical Angle
Question:
In figure, five light rays enter a glass prism from the left.
(i) How many of these rays undergo total internal
reflection at the slanted surface of the prism? (a) one
(b) two (c) three (d) four (e) five
Example:
Find the critical angle at an interface between glass (n = 1.50) and air.

Example:
You are told that a new optical material has a critical angle in air of 34.4°. What is
the index of refraction of this material?
Since a critical angle is defined, the index of the material must be higher than air.

82
Critical Angle
Example:
A particular glass has an index of refraction of n = 1.50. What is the critical angle for
total internal reflection for light leaving this glass and entering air, for which n = 1.00?

Example:
What is the critical angle for light traveling in a polystyrene (a type of plastic) pipe
surrounded by air?
The index of refraction for polystyrene is found to be 1.49 and the index of refraction
of air can be taken to be 1.00, as before. Thus, the condition that the second
medium (air) has an index of refraction less than the first (plastic) is satisfied, and
the equation θc = sin−1(n2/n1) can be used to find the critical angle θc . Here, then, n2
= 1.00 and n1 = 1.49. Substituting the identified values gives
83
Critical Angle
Example:
In the given figure; at point B,at least part of the light reflects and leaves the
diamond properly, but part could refract and thus leak out of the diamond. Consider
a light ray incident at angle fu θ1 = 40 at A. Does light Ieak at B if air (n4 = 1.00) lies
next to the bottom surface? Does light leak if greasy grime (n4 = I.63) coats the
surface? The index of refraction of diamond is ndia : 2.419.




84
Reduced Distance
• When an object in one medium is viewed from another medium, the apparent
position of the object differs from its actual position.
• The apparent image displacement in water appear to be closer than they actually
which is just another consequence of refraction.
• i.e. A fish is located at a distance t below
the surface of the water.
• The reduced distance -or equivalent
thickness or apparent position- to the
fish is the distance d, and is equal to
• Objects in water seem less deep than
they really are.
FIGURE: The fish is located at a distance t
from the surface of the water. It appears to be
located at the reduced distance d, however,
because of the refraction of light.
85
Reduced Distance
Example:
A person stands above the water looking at an object located 1 m below the surface
of a pond (n = 1.333). How far from the surface does the object appear to be located
(what is the reduced distance)?
d = 1/1.333  d = 0.75
∴ Reduced distance is 0.75 m
Example:
A pebble located at the bottom of a fish tank appears to be 22.55 cm from the
surface. What is the depth of the water in the tank?
Use the general rules for the variables:
n = 1.33 (object in water);
n' = 1.00 (viewing from air);
l= ? (actual distance of object from interface is
the depth of the water);
l’ = 22.55 cm (apparent position of object from
interface).

The apparent position is 7.45 cm (30.00 cm - 22.55 cm = 7.45 cm) closer than the
actual pebble.
86
Reduced Distance
Example:
You are enjoying a n ice break at the pool. While under the water, you look up and
not ice that you see objects above water level in a circle of light of radius
approximately 2.0 m, and the rest of your vision is the color of the sides of the pool.
How deep are you in the pool?


87

Reduced Distance
Example:
A small fish is swimming at a depth d below the surface of a pond (see figure).
(a) What is the apparent depth of the fish as viewed from directly overhead?
Figure: (a) The apparent depth q of the fish is less than the true depth d. All rays are assumed to be
paraxial. (b) Your face appears to the fish to be higher above the surface than it is.
88
Reduced Distance
Example (cont.):
(b) If your face is a distance d above the water surface, at what apparent distance
above the surface does the fish see your face?
(c) What if you look more carefully at the fish and measure its apparent height from
its upper fin to its lower fin? Is the apparent height h’ of the fish different from the
actual height h?
The apparent height h’ of the fish is
Hence, the fish appears to be approximately three-fourths its actual height.
89
Reduced Distance
Example:
A small fish is swimming at a depth d below the surface of a pond (see figure).
What is the apparent depth of the fish, as viewed from directly overhead?
Because the refracting surface is flat, R is infinite. Hence,


Because q is negative, the image is virtual.
90
Reduced Distance
Example:
A fish in the tank views a fly that appears to be 7 cm above the water surface. How
far above the surface is the fly?

Note that the actual position of the fly is
closer than the apparent position in this
case.
• From these examples, general rules about apparent position may be developed:
91
Apparent Position: Multiple Parallel-Sided Surfaces
• Reduced distance is the thickness of a parallel sided plate or the path a ray
travels divided by the index of the medium in which it travels:
• The apparent position of an object viewed through multiple parallel-sided
interfaces may be calculated by adding the reduced distance of each interface.
• The final position is determined relative to the last interface.
• This is written as
• where n' is the index from which the object is viewed, and l is the apparent
position relative to the last interface.
92
Apparent Position: Multiple Parallel-Sided Surfaces
Example:
In below figure, an object is located under a tank filled with three media. The
indices and thickness of the media are labeled. Where is the apparent position of
the object viewed from the air above the tank?
Solve for the apparent position using Equation
In this case, the apparent position of the object is
6.56 cm from the last interface of oil-air. The
actual position of the object is 11 cm from the last
interface. The apparent position is significantly
closer than the actual object.
93
Apparent Longitudinal Displacement
• Apparent longitudinal displacement is the apparent position of an object viewed
normally through a parallel-sided plate surrounded by the same medium.
• The displacement may be calculated using this equation:
• In this equation, t is the thickness of
the plate, n' is the index of the plate,
n is the index of the surrounding
medium, and d is the apparent
displacement of the object.
• The displacement d is relative to the
actual position of the object, not the
interface. (see below figure).
• Apparent thickness is the apparent
depth or thickness of a parallelsided plate viewed normal to its
surface.
Figure: Displacement of object viewed through a
parallel plate.
94
Apparent Longitudinal Displacement
Example:
An object is viewed through a 5cm thick glass plate (n = 1.45) normal to the surface.
The object is 20 cm from the front of the plate. Where is the apparent position of the
object relative to the actual object position and to the back surface of the glass
plate? What is the apparent thickness of the glass plate?
The glass plate with index n' is surrounded by air (n = 1.00), and the displacement
along the axis is calculated using as
The apparent position of the object is 1.55 cm from the actual object. In this
example, the surrounding medium has a lower index than the parallel-sided plate,
and therefore the apparent position is closer to the plate than the actual object. The
apparent position of the object relative to the back of the plate may be
calculated by:
distance to object from front of plate - thickness of plate - displacement
20 cm - 5 cm - 1.55 cm = 13.45 cm
The apparent thickness of the plate may be calculated as


95
Lateral Displacement
• Lateral Displacement is the perpendicular distance between an incident and
emerging ray after traveling through parallel sided interfaces.
• See below figure, where d is the lateral displacement.
Example:
Show the relationship between the thickness of a parallel-sided plate, the incident
and refracted angles, the indices, and the lateral displacement. If a glass plate (n =
1.56) is 15 cm thick, what is the lateral displacement for a ray incident at 45°?
Figure: Parallel-sided plate with the lateral
displacement labeled d.
Figure: Triangles required for the derivation
of lateral displacement.
96
Lateral Displacement
Example (Cont.):
The angle that encompasses both triangles is labeled θ1, the refracted angle in the
lower striped triangle is labeled θ’1 , and the angle in the shaded triangle is the
difference between these two angles or {θ1 – θ’1} as indicated. The tangent of this
angle is defined as
where ab is the hypotenuse of the triangle, and d is the lateral displacement.
From the striped triangle, calculate the value ab using the cosine function:
where: t = the thickness of the plate. Substituting for ab in above equation and
solving for d, this becomes

97
Lateral Displacement
Example (Cont.):
For the incident angle of 45°, the refracted angle must be determined using Snell's
Law:


Using abobe equation, solve for the displacement:

Question:
Show that a ray will emerge from air-glass-water interfaces at the same angle as a
ray that emerges from a single air-water interface.
Figure: The emerging angle (18.53°) will be the same in both cases above.
98
INTERFERENCE FROM THIN FILMS
• When oil floating on water is viewed in sunlight, beautiful colours appear.
• These colours appear because of interference between the light waves sent by
the film as explained below.
• Consider a thin film made of a transparent material with plane parallel faces
separated by a distance d.
• Suppose a parallel beam of light is incident on the film at an angle i as shown in
below figure.
99
INTERFERENCE FROM THIN FILMS
• The wave is divided into two parts at the upper surface, one is reflected and the
other is refracted.
• The refracted part, which enters into the film, again gets divided at the lower
surface into two parts; one is transmitted out of the film and the other is reflected
back.
• Multiple reflections and refractions take place and a number of reflected waves
as well as transmitted waves are sent by the film.
• The film may be viewed by the reflected light (more usual case) or by the
transmitted light.
• We shall discuss the transmitted light first.
• The optical path difference between the waves BP and DP interfering at P is
Δx = 2nd
• The phase difference is
 = 2Δx/ = (2)(2nd/)
• This is also the phase difference between the waves DP and FP or in fact,
between any consecutively transmitted waves.
• All these waves are in phase if
 = 2m or 2nd = m
….. (1)
• where m is an integer.
100
INTERFERENCE FROM THIN FILMS
• If this condition is satisfied, constructive interference takes place and the film is
seen illuminated.
• On the other hand, if
2nd = (m +1/2)  ….. (2)
•  = (2m + 1) and the consecutive waves are out of phase.
• The waves cancel each other although complete cancellation does not take
place because the interfering waves do not have equal amplitude.
• If white light is used, the film's thickness d will satisfy condition (1) for certain
wavelengths and these colours will be strongly transmitted due to constructive
interference.
• The colours corresponding to the wavelengths for which (2) is satisfied will be
poorly transmitted due to destructive interference.
• This gives coloured appearance of the film.
101
INTERFERENCE FROM THIN FILMS
• Next, let us consider the case when the film is viewed by the light reflected by it.
• In this case, the conditions for maximum and minimum illumination in reflection
should be opposite to that in transmission.
• We should have
2nd = m for minimum illumination in reflection ….. (3)
and
2nd = (m +1/2)  for maximum illumibation in reflection ….. (2)
Example:
Find the minimum thickness of a film which will strongly reflect the light of
wavelength 589 nm. The refractive index of the material of the film is 1.25.
Solution : For strong reflection, the least optical path difference introduced by the
film should be /2. The optical path difference between the waves reflected from
the two surfaces of the film is 2nd. Thus, for strong reflection,
2nd = /2  d = /4n = 589 nm / (4 x 1.25) = 118 nm
102
INTERFERENCE FROM THIN FILMS
Example:
You observe colored rings on a puddle and conclude that there must be an oil slick
on the water. You look directly down on the puddle and see a yellow–green ( =
555 nm) region. If the refractive index of oil is 1.45 and that of water is 1.33, what
is the minimum thickness of oil that could cause this color?
Because noil > nair, there is a phase inversion
on the first reflection. Because nwater < noil,
there is no phase inversion on the second
reflection. Thus, there is one wave inversion.
The wavelength in oil is less than it is in air.
Follow the problem-solving strategy to
construct the equation.
Because you want the
minimum thickness, m = 0.
103
Questions
1. A bird is 30 feet above a swimming pool. You are submerged and looking
straight up at the bird. How high above the surface of the water does it appear
to be? Ans.: 40 ft.
2. You are 6 ft under water. How far under water do you appear to be to the bird?
Ans.: 4.5 ft.
3. A shark in an aquarium tank longingly looks at the visitors. The shark spots you
at eye level. You appear to the shark to be 5 ft from the tank wall. How far from
the wall are you? Ans.: 3.75 ft
4. The shark appears to you to be 16 inches from the glass tank wall. How far is
it? Ans.: 21.33 in.
5. A cube of 1.6 index glass, 80 mm on a side, is used as a paper weight. By how
much does the paper appear to be raised off the table when you look straight
down? Ans.: 30 mm.
6. A beaker contains layers of immiscible liquids of the following thicknesses 10,
20 and 30 mm, with respective indices of 1.3, 1.4 and 1.45. How deep do the
liquids appear to be, on looking down into the beaker? Ans.:
7.69+14.29+20.69=42.67 mm.
104
Questions
1. A bank teller sits 18 inches behind a 3-inch thick Plexiglas window of index 1.5.
You are 12 inches from the near side of the window. How far from you does the
teller appear to be? Ans.: -32 in.
2. A microscope is focused on a table top. A parallel plate of glass 24 mm thick is
placed on the table and the microscope must be raised 10 mm, to refocus on the
top of the table. What is the index of the glass? Ans.: 1.714
3. A 24-cm thick glass plate of index 1.5 divides a tank, filled with water of index
4/3, into two sections. A submerged swimmer in each section is 24 cm from the
nearer face of the plate. How far apart from each other do they appear to be
now? Ans.: 69.333 cm
4. A silicon solar cell has a non-reflective coating placed on it. If a film of sodium
monoxide, n = 1.45, is placed on the silicon, n = 3.5, how thick should the layer
be to keep yellow green light ( = 555 nm) from being reflected?
5. You can observe thin-film interference by dipping a bubble wand into some
bubble solution and holding the wand in the air. What is the thickness of the
thinnest soap film at which you would see a black stripe if the light illuminating
the film has a wavelength of 521 nm? Use n = 1.33.
6. What is the thinnest soap film (n =1.33) for which light of wavelength 521 nm will
constructively interfere with itself?
105
Questions
1. The velocity of light in sea water is 224,551 km/sec. What is the index of
refraction of the water? Ans.: 1.336
2. Light is incident at an angle of 30° at a plane refracting surface separating air
from glass of index 1.5. Find: a) the angle of refraction? (b) the angle of
deviation? Construct the refracted ray. Ans.: a. 19.471°; b. 10.529°
3. Light is incident at an angle of 30° at a plane refracting surface separating glass
of index 1.6 from water of index 1.333. Find: (a) the angle of refraction; (b) the
angle of deviation; the critical angle; d) the relative index of refraction. Construct
the refracted ray. Ans.: (a) 36.881°; (b) -6.881°; (c) 56.421°; d) 0.833
4. The angles of incidence and refraction at a plane interface, separating air from a
second medium, are 50° and 35°, respectively. The velocity of the light in air is
300,000 km/sec. Find (a) the relative index; (b) the absolute index of the second
medium; (c) find the velocity of the light in the second medium; d) find the
relative index when the media are reversed. Construct the ray path. Ans.: (a)
1.3356; (b) 1.3356; (c) 224,625 km/sec; d) 0.7488.
5. A ray makes an angle of incidence of 30° with a plane interface that separates
water (index = 1.3333) from glass (index = 1.7500). Find: (a) the angle of
refraction; (b) the angle of deviation; (c) the critical angle. Construct the ray path.
Ans.: (a) 22.3927°; (b) 7.6073°; (c) 49.6324°
106
Questions
1. A ray makes an angle of incidence of 45° with a plane interface that separates
glass (index = 1.7500) from glass (index = 1.5000). Find: (a) the angle of
refraction; (b) the angle of deviation; (c) the maximum angle of incidence that
results in a refracted ray; d) the angle of refraction of this ray. Construct the ray
path. Ans.: (a) 55.5842°; (b) -10.5842°; (c) 58.9973°; d) 90°.
2. A beaker contains a 1 inch thick layer of water (index 1.333) A 0.5 inch layer of
oil (index 1.45) floats on the water. A ray of light makes an angle of incidence of
60° with the upper surface of the oil. Find the angles of refraction at the air/oil
and oil/water interfaces. Ans.: 36.674°; 40.518°
3. What is (a) the actual path length and b) the optical path length of the ray from
the point of incidence at the oil to the bottom of the beaker? Ans.: a. 1.938 in.; b.
2.657 in.
4. You are 3 feet under water and looking straight up. Within what total angle will
the heavens be contained? Ans.: 97.18°
5. In previous problem 4, what is the diameter on the water's surface through
which light from the sky reaches your eye? Ans.: 6.8 ft.
6. White light is incident at an angle of 45° at a plane refracting surface separating
air from glass. The index of refraction of the glass for red light is 1.500; for blue
light it is 1.505. (a) What is the angle between the refracted red and blue rays?
(Note: this is called angular dispersion, b) which color is deviated more? Ans.:
(a) 6.1 minutes of arc. (b) blue.
107
Image-Object Relationships: Size and Orientation
• Just as the positions of the images and objects are related, their relative sizes
and orientation (right side up or upside down) are also related.
• Below figure illustrates a positive single refracting surface. The object and image
are shown with respective sizes of h and h’.
• The relative size of the image and object can be expressed as a ratio called the
lateral magnification:
FIGURE: Radius method for determining lateral magnification.
108
Image-Object Relationships: Size and Orientation
• This ratio is greater than unity if the image is larger than the object and it is less
than unity when the image is smaller than the object.
• The sign associated with the lateral magnification indicates the image-object
orientation.
• If the image and object are on opposite sides of the axis (one above and the other
below), the relationship is called inverted, and the lateral magnification is
negative.
• If the image and object are on the same side of the axis (both above or both
below), the relationship is called erect, and the lateral magnification is positive.
• Using the two shaded similar triangles shown in the above figure, the following
relationship may be developed:
• Setting the equation equal to the lateral magnification (h'/h), this becomes
109
Image-Object Relationships: Size and Orientation
Figure: Illustration of Snell's Law to determine lateral magnification.
• Using Snell's Law
• Because the region about the axis is considered to be paraxial, the sine of the
angle is equal to the tangent of the angle, which is equal to the angle itself in
radians.
• This may be expressed as:



110
Image-Object Relationships: Size and Orientation
FIGURE: Extrafocal distance method for determining the lateral magnification.
• The lateral magnification may be determined by using the shaded triangles.
• The shaded triangles are similar, and the relationship may be written as
• The minus sign is included in the equation so that the sign convention can be
consistant.
111
Image-Object Relationships: Size and Orientation
• In here; x is the extrafocal object distance which is the distance from the primary
focal point to the object and x’ is the extrafocal image distance which is the
distance from the secondary focal point to the image.
• The relationship between the extrafocal distances, focal lengths, and image and
object distances may be seen in
• Another lateral magnification relationship may be developed by using the similar
striped triangles on the right:
• The extrafocal distances may also be used to calculate the lateral magnification
of the thin lens.
• Note that the lateral magnification may also be determined by using the ray tracing
procedure.
112
Image-Object Relationships: Size and Orientation
• The symmetrical planes are the image and object distances that yield a lateral
magnification of - 1X.
• This means that the object and image are the same size but inverted.
• This relationship may be written as
• The minimum distance between the symmetrical planes is the difference
between the image and object positions.
• This may be written as:
113
Image-Object Relationships: Size and Orientation
• When an object is at an infinite distance, the lateral magnification cannot be
defined since l is infinite
FIGURE: Image size with an infinite object.
• The image in the secondary focal plane may be drawn with the proper height.
• The shaded triangle is used to calculate the height of the image:

• For small angles; this may be written as
114
Examples on Refraction of Light
Example: Converging Surface
An object is located 50.00 cm to the left of a +5.00 D spherical crown glass surface.
How far is the image from the apex of the surface? Is the image real or virtual? Is it
erect or inverted? If the object is 3.00 cm in height, what is the height of the image?
• The object vergence at the refracting surface is
L = n/l  L =1.00/−0.50 m  L = = −2.00 D
• Light with a vergence of −2.00 D is incident upon a surface whose power is +5.00 D.
115
Examples on Refraction of Light
• Example (Cont.):
• According to the vergence relationship, the vergence of the light rays that form the
image is the sum of the incident vergence and the surface power:
L’ = L+ F

L’ = -2.00 D + (+5.00 D) 
L’ = +3.00 D
• Since converging light rays from the image, it is real.
l′ = n′ / L′

l′ = 1.52 / +3.00 D
 l′ = +0.5067 m, or +50.67 cm
• The image is located 50.67cm from the surface.
• The image size is calculated from lateral magnification (ML) which is equal to the
ratio of the object vergence to the image vergence as indicated below:


• Since the object height is 3.00 cm, the image height is
(−0.66) x (3.00 cm) = −1.98 cm
• Because the magnification is less than 1.00x, the real image is smaller than the
object, it is minified.
• Negative magnification tells us that the image is inverted.
116
Examples on Refraction of Light
Example-2:
A +10.00 D lens is used to view the printed words on a page located 9.00 cm from the
lens. Is the image formed by the lens real or virtual? Is it erect (dik) or inverted?
Where is it located, and what is the lateral magnification?
• The object vergence is
 L = -11.11 D
• The paraxial relationship gives us L′ = L+F  L′ =−11.11 D+10.00 D  L′ =− 1.11 D
• The image distance is
 l ′ = −90.10 cm
• The virtual image is located 90.10 cm to the left of the thin lens.
117
Examples on Refraction of Light
Example-3:
A 7.00 mm high object is located 8.00 cm from a +15.00 D lens. Is the image real or
virtual? Is it erect or inverted? Where is it located, and what is its size?
• The object vergence
L = n / l  L = −12.50 D
• To determine the image
vergence, we used the
vergence relationship
L′ = L + F 
L′ = −12.50 D + 15.00 D  L′ = +2.50 D
• Since the vergence is positive, we know that the image is located to the right of the
lens, real, and inverted. Its location is
L′ = n′/l ′  l′ = + 40.00 cm
• The magnification is M = l ′/l  M = 40.00 cm / −8.00 cm  M = −5.00×
• The object height is 7.00 mm, making the image height
(−5.00)(7.00 mm) = −35.00 mm
• The minus sign reminds us that the magnified image is inverted.
118
Examples on Refraction of Light
Example-4: In below figure, rays emitted from an axial object hit a wavefront
changer (lens) and are directed toward an axial image. The distances from the
lens to the object and the lens to the image are labeled. Find the radius of the
incident and emerging wavefronts at the lens, and calculate the vergence for the
respective pencils. How much did the lens change the vergence?
The object vergence is calculated by
L = 1 / l  L = 1/-0.2  L = −5.00 D
The image vergence is calculated by
L′ = 1 / l ′  L’ =1 / +0.4  L’ = +2.50 D
An incident vergence of - 5.00 D changes
to a vergence of + 2.50 D after passing
through the lens. Mathematically, this can be written as
Incident (object) vergence + Lens power = Emergent (image) vergence
Substituting the vergence into the formula gives
Lens power = + 2.50 D - (- 5.00 D) = + 7.50 D
This means a + 7.50 D lens is required to change the incident divergence of - 5.00
D to the emergent convergence of + 2.50 D.
119
Examples on Refraction of Light
Example-5:
Light is incident onto a wavefront changer with a vergence of + 3.00 D. The light
emerges from the wavefront changer with a vergence of - 4.00 D. Where are the
object and image located? What is the power of the wavefront changer?
The incident light is a convergent pencil with a vergence of + 3.00 D.
L = 1 / l  l = 1/+3.00  l = +0.33 m = +33.33 cm
The object location is therefore 33.33 cm to the right of the wavefront changer
because the radius is positive. This will be referred to as a virtual object because it
is formed with converging light.
120
Examples on Refraction of Light
Example-5 (Cont):
The emerging pencil is divergent (as indicated by the negative vergence), and the
origin of the wavefront can be located by solving for the radius.
L‘ = 1 / l’  l’ = 1/-4.00  l’ = -0.25 m =-25.00 cm
The origin of the emerging wavefront is the image; the negative value indicates that
its position is to the left of the wavefront change.
The power of the wavefront changer is calculated with the incident and emerging
vergence and the power of the wavefront changer (lens) yields
Incident (object) vergence + Lens power = Emergent (image) vergence
Lens power = L’ - L = - 4.00 D - (+ 3.00 D) = - 7.00 D
121
Examples on Refraction of Light
Example:
A pencil of rays leaves a lens and converges towards a point 140 cm from the lens
as shown in below figure. What is the vergence at a point 60 cm to the right of the
lens?
The distance from the point at which the
vergence is to be determined to the center
of the wavefront is given by
140 cm - 60 cm =80 cm
The radius of the wavefront is + 80 cm
because it is measured from left to right
(in the direction light travels) and the
vergence is
L′ = 1 / l ′  L’ =1 / 0.80 m  L’ = +1.25 D
122
Examples on Refraction of Light
Example:
An object was placed 1 m in front of a + 5.00 D single refracting surface, and a
real image was located 37.50 cm from the surface. The index of object space was
1.00, and the index of image space was 1.50. The radius of the surface is 10 cm.
Solve for the lateral magnification using each of the techniques discussed. Is the
final image inverted or erect?
Solve for the lateral magnification using the following equations
To calculate the focal lenghts and extrafocal distances, the following equation are
respectively used:
123
Examples on Refraction of Light
Example (Cont):
Solving for the lateral magnification using the following equations:
124
Examples on Refraction of Light
Example:
A - 3.00 D lens forms an image that is erect and five times larger than the object.
Where is the object located?
Set up the lateral magnification relationship using the following equation:


125
Examples on Refraction of Light
Example:
Locate the symmetrical planes for a + 5.00 D lens and a - 5.00 D lens.
126
Examples on Refraction of Light
Example:
The angular subtense of an object is often used in vision to describe the size of a
distant object. For example, a 20/60 letter on an eye chart subtends a 15 minute
arc at 20 feet (1 degree = 60 minutes). If a + 10.00 D lens is used to form a distant
letter on a screen, what is the size of the image formed:
First, solve for the secondary focal length of the lens using the following equation:
This represents the image plane location. Next, convert the minutes of arc to
degrees:
Using this value, solve the following equation for the height of the image:
The negative sign indicates that the image is inverted. A small image is formed by
the lens using the 20/60 letter as the object. The eye actually resolves images
that have critical dimensions (1/5 the size of the letter) on the order of 0.75 min
(20/15 acuity).
127
POWER of MIRRORS
• As mentioned previously, when light passing across a curved interface between
two media of different refractive indices obeys Snell's law.
• A convex spherical curved surface causes parallel light
• to converge to a focus if n2 is greater than n1,
• or to diverge as from a point focus if n2 is less than n1.
• The refracting power or
vergence power of such a
surface is given by the
formula
• where r is the radius of
curvature of the surface in
metres according to the
sign convention and the
surface power is measured
in dioptres.
Figure: Refraction of light at a convex refracting interface, e.g. cornea.
128
Power of Mirrors
• Surface power:
• is positive for converging surfaces
• and negative in sign for diverging surfaces.
• The relationship between the power of a mirror and the secondary focal length is
• If the mirror is in air, then
Example:
A concave mirror has a radius of curvature of 10.00 cm. Where is the focal point
located with respect to the mirror’s surface? What is the mirror’s power?
• The focal length is calculated as follows:
r = 2f ′  −10.00 cm = 2f ′  f ′ = −5.00 cm
• The focal point is located 5.00 cm to the left of the mirror’s surface.
• The mirror’s power is
F = −1/f ′  F = −1/−0.05 m  F = +20.00 D
129
LENS POWER
• The curvature (R) of a surface is inversely proportional to its radius (r) of
curvature; as the radius increases, the curvature decreases (R =1/r).
• Hence, the surface power (F) is directly proportional to both the amount of
curvature (R) and the refractive index (n) of the lens material.
• An increase in either the magnitude of the curvature, or the refractive index, will
result in an increase in the magnitude of the surface power.
• The ophthalmic lenses are described in terms of refracting power and it is defined
as the change in vergence that occurs when light passes through a lens.
130
Lens (Refractive) Power
• There is a number of ways of specifying the refracting power of an ophthalmic lens:
• Approximate power FA (also called nominal power) according to which the
power of a lens is specified in terms of its front and back surface powers without
regard to thickness;
• Back vertex power and front vertex power according to which the power of a
lens is specified in terms of the refracting power for emergent rays at its back
surface or front surface;
• Equivalent power according to which the power of a "thick" lens or optical
system is specified in terms of the power of a single thin lens;
• Effective power FE according to which the power of a lens is dependent upon its
distance from the wearer's eye.
• Of all of these methods of power specification, only back vertex power is used
routinely by optical laboratories.
• Instrumentation for the measurement of back vertex power includes the lensometer
or vertometer, the projection lensometer, and the automatic (computerized)
lensometer.
131
Lens Power
• In a lens with a positive power, the light rays converge or are refracted toward one
another.
• The point at which the light rays converge is called the focal point and it is behind
the lens surface.
• In a lens with a negative power, the light rays diverge or are refracted away from
one another.
• If these rays are extrapolated or traced back toward the light source, the lines will
converge and form a focal point in front of the lens surface.
• The lens power (vergence power) is the reciprocal of the focal distance in meters
and it is expressed in diopters (D).
• Note: 1 diopter of focal power will focus light at a distance of 1 meter.
Example:
Light rays pass through a lens and converge 0.50 m from the lens. What is the
power of the lens?
Power = 1 / Focal Distance  Power = 1 / 0.50 m  Power = 2.00 D
132
Near and Far Points
• For eye:
• i-) The near point is the closest distance
for which the lens can accommodate to
focus light on the retina.
• This distance changes as the eye ages:
Age 10: near point = 18 cm.
Age 20: near point = 25 cm.
Age 40: near point = 50 cm.
Age 60: near point = 500 cm.
TABLE: Some common diopter and focal length
equivalents.
• ii-) The far point is the farthest distance
for which the lens of a relaxed eye can
focus light on the retina.
• People with normal vision can focus
objects that are very far away.
133
Interface
• Interface is the boundary (plane, curved, aspheric, etc.) between two media with
different indices.
• In below figure, light travels in medium 1 (left) before striking the interface of
medium 2 (right).
• Usually n is the label for the index of the medium which the incident ray travels in
before striking the interface.
• This is the first medium listed when describing an interface.
• The other medium of the interface is labeled n' and is the second media listed.
• Below figure presents several plane or flat interfaces with the proper labels.
Figure: Examples of interfaces between two media.
134
Types of Reflections
• There are two principal types of reflections that can occur.
• When rays of light strike a smooth and shiny surface, like glass, the surface
reflects an image of the source.
• This is known as specular reflection where the reflected rays are all parallel to
each other.
• In this case, the angle of incidence is equal to the angle of reflection.
• A surface behaves as a smooth surface as long as its variations are small
compared with the wavelength of the incident light.
• When rays of light strike a rough surface, like concrete, the uneven surface
reflects, or scatters, the light in every direction.
• This is known as diffuse reflection where the reflected rays travel in random
directions.
• This occurs at irregular surfaces, such as a projection screen or frosted glass.
FIGURE: Rough (diffuse) versus smooth (specular) surface reflection.
135
Reflection
• As an example, consider the two types of reflection from a road surface that a
driver might observe while driving at night.
• When the road is dry, light from oncoming vehicles is scattered off the road in
different directions (diffuse reflection) and the road is clearly visible.
• On a rainy night when the road is wet, the road’s irregularities are filled with water.
• Because the wet surface is smooth, the light undergoes specular reflection.
• This means that the light is reflected straight ahead, and the driver of a car sees
only what is directly in front of him.
• Light from the side never reaches his eye. In this book we concern ourselves only
with specular reflection, and we use the term reflection to mean specular reflection.
Figures: Photographs of specular and diffuse reflection, made with laser light.
136
Reflection
Question:
You may have noticed a common occurrence in photographs of individuals: their
eyes appear to be glowing red. Why?
This occurs when light from the flash unit enters the eye and is reflected back along
its original path from the retina. This type of reflection back along the original
direction is called retroreflection. If the flash unit and lens are close together,
retroreflected light can enter the lens. Most of the light reflected from the retina is
red, due to the blood vessels at the back of the eye, giving the redeye effect in the
photograph.
Question:
Which part of the below figure (a) or (b), better shows specular reflection of light
from the roadway?
137
Reflection
• The reflected light from spectacle lenses (gözlük
troublesome to the wearer.
camları)
surfaces may be
• For example, it can produce:
• ghost images (hayalet),
• Falsification (taklit, değiştirme) of image position,
• Haze (bulanık, sis),
• and loss of contrast.
Example:
When looking outdoors through a glass window at night, why do you sometimes see
a double image of yourself ?
Reflection occurs whenever there is an interface between two different media. For
the glass in the window, there are two such surfaces. The first is the inner surface of
the glass, and the second is the outer surface. Each of these interfaces results in an
image.
138
REFLECTION
• The light passing through any material (i.e. a lens) may be either
• completely transmitted if the lens is perfectly clear,
• or absorbed to some extent if the lens has filtering properties,
• or reflected from the material.
• When light reaches the boundary or interference between two transparent media
(i.e air and water) having different indices (i.e. na = 1 and nw = 1.33) of refraction,
most of the light is refracted but a small amount is reflected.
• For example, when light is incident upon a lens, some of it is reflected by each of
the lens surfaces, some of it is absorbed by the lens, and the remainder is
transmitted.
• The percentage of the incident light that
leaves the lens after passing through it (the
transmitted light) is found by calculating
• the percentage light lost by reflection at the
front surface,
• the percentage lost by absorption,
• and the percentage lost by reflection at the
back surface.
FIGURE: Reflection, absorption, and
transmission by a lens
139
Reflection
• In the reflection process; the amount and color of light that is reflected back into
the first medium depends upon
• the nature of the materials (i.e. refractive index of medium, color of the
materials (black absorbs and white reflects), surface of the materials (smooth,
polished, rough, etc).
• as well as the angle of incidence.
• All transparent materials reflect a certain amount of light.
• The fraction of incident light reflected from a surface is referred to as the
reflectance ρ.
• Reflectance is calculated from the refractive index of a material.
• As the refractive index of a surface increases, the amount of light that is reflected
increases.
140
Fresnel's Law of Reflection and Transmission
• For light rays that are perpendicular to a transparent surface, the fraction of light R
= IR /I that is reflected is given by the following relationship (Fresnel’s formula):

• where n is the refractive index of the medium surrounding the surface (generally
air) and n′ is the refractive index of the surface.
• Similarly, the fraction that is transmitted at a single interface can be found by
applying the Fresnel equation, which for normal incidence is
• İf we assume the only losses are by reflection, the transmittance T(k), for a
system of & identical surfaces, is given by the equation
141
Fresnel's Law of Reflection and Transmission
Example:
How much light is reflected at air-water, water-air, and water-crown glass
interfaces? How do the indices of the media that form the interfaces affect the
amount of reflection? Assume the incident ray is normal to the interface.
Using Fresnel's Law of Reflection, the amount of reflected light is calculated for
each interface:
From the calculations, the order of the media (air-water or water-air) does not affect
the amount of reflected light. The closer the indices of the media that form the
interface, the lower the amount of reflected light.
Question: Calculate the approximate reflectance of the moon's surface if it is
perfectly diffusing.
ANSWER: 0.069
142
Fresnel's Law of Reflection and Transmission
Example:
Calculate the surface reflection at an air/glass interface where the glass is Schott
glass BK 7, for a single lens and a system of 10 identical surfaces.
Solution: From the Schott (1992) glass catalog, the refractive index of BK 7 glass
in the middle of the visible spectrum is very close to 1.5187. Substituting n = 1 and
n' =1.5187 into below equations we have


For a lens, there are two air/glass surfaces and so the fraction of light transmitted
is, from below equation,
For an optical system with 10 air/glass surfaces, the fraction of light transmitted is
Thus in this case, about 35% of the light is lost.
143
Fresnel's Law of Reflection
Example:
According to Fresnel's Law of Reflection, how much light is transmitted through a
piece of window glass (index of refraction = 1.50). Assume the thickness of the
glass is negligible and that no light is absorbed.
To determine the total amount of light transmitted through the glass, one must first
determine the amount of light lost by reflection at each surface (see figure). The
first surface is an interface between air and glass, and the amount of light reflected
is given by
If we consider the incident light to be
100% and 4% of this light to be reflected,
then the transmitted light at the first
surface is: 100% - 4% = 96%
The amount of light reflected at the second surface (glass-air) is the same as the
amount of light reflected at the first surface (air-glass) or 4% of the incident light.
Since 4% of the 96% or 3.84% (0.96% x 0.04 = 0.038 ) of the incident light is
reflected at the second surface, the amount of light transmitted is
96%-3.8% = 92.16%
Approximately 92% of the incident light is transmitted through the glass.
144
Fresnel's Law of Reflection
• By inspection of Fresnel's equation, we may conclude the following:
1. The intensity (IR) of the reflected beam is unaffected by the direction of the
incident light, that is, whether it traverses from air to glass or from glass to air.
2. The intensity of the reflected beam increases as the index of refraction
increases. Hence, the reflection intensity is greater for lenses made with high
index materials. The percentage reflected at an air-water (n = 1.33) interface
is 2.0%, while the percentage reflected at an air-diamond (n = 2.4) interface is
17.0%. For CR-39 (n = 1.498), a hard resin plastic used for spectacle lens
corrections, the percent reflection in air is 4.0%. For highlite glass (n = 1.71),
a high index glass used for spectacle corrections, the percent reflected in air
is 6.9%.
3. Since the index of refraction increases toward the blue end of the spectrum,
the reflected image will contain more blue light than red light. Therefore the
reflected image may appear slightly bluish in color while the transmitted light,
deficient in blue, may appear slightly yellowish.
145
Reflection
Example: Light is incident upon a transparent medium with an index of refraction of
1.500. What is the reflectance?
When light hits a lens surface in air normally, the percentage of light reflected at
each surface is given by:


ρ = 0.04 ∴ Reflectance is 0.04 (or 4%).
Thus a material of refractive index 1.5, has a reflectance of 4% per surface
When thr index of refraction of the lens material increases the amount of light reflected
increases.
Some reflection also occurs at every
interface even though in this case most of
the incident light passes onwards by
refraction.
For example, a lens or window with a
refractive index of 1.5 in air reflects 4% of
light from the anterior surface and transmits
the remaining 96% to the posterior surface;
a further 4% of this is reflected so that the
lens transmits only 92.16% of normally
incident light
Thin coatings can be applied to a lens surface to reduce this reflectance to almost
nothing—thereby increasing the transmittance of the lens.
146
Reflection
Example:
• For clear ophthalmic crown glass, having an index of refraction of 1.523, the light
lost by reflection at the front surface is
• Since clear ophthalmic crown glass absorbs no light, the percentage of incident
light remaining when the light reaches the back surface of the lens is
• and the light lost by reflection at the back surface of the lens is
• and the light transmitted by the lens is
Question: For a plastic lens (CR-39 material) having an index of refraction of 1.498,
Calculate (a)- the light loss at the front surface, (b)- the light loss at the back surface,
(c )- light transmitted by the lens.
147
Reflection
Example:
Light in air is paraxially incident on an air-ophthalmic crown glass (n = 1.523)
interface. What percentage of the incident light is reflected?


4.3% of the incident light is reflected at the air-glass interface. Therefore, 95.7% of
the incident light is transmitted through the interface and refracted. The percent
reflected is independent of which medium the light is initially in. So if the light is
reversed and is incident on the air-glass interface from the glass side, then again
4.3% of the incident light is reflected.
EXAMPLE:
Light is paraxially incident on a water (n = 1.33)-plastic (n = 1.49) interface. What
percent of the incident light is reflected?

Only 0.3% of the incident light is reflected at the water-glass interface, so 99.7% is
transmitted.
148
Brewster Angle
• For light incident at the Brewster angle θB, we
find experimentally that the reflected and
refracted rays are perpendicular to each other.
Because the reflected ray is reflected at the
angle θB in below figure and the refracted ray is
at an angle θr, we have


• If the incident and reflected rays travel in air, we can approximate n1 as unity
and let n represent n2 in order to write equation as
149
Brewster Angle
Question-1: At what angle of incidence will the light reflected from water be
completely polarized? (b) Does this angle depend on the wavelength of the light?
Question-2: Light that is traveling in water (with an index of refraction of 1.33) is
incident on a plate of glass (with index of refraction 1.53). At what angle of incidence
does the reflected light end up fully polarized?
Question-3: In below figure, a light ray in air is incident on a flat layer of material 2
that has an index of refraction n2 = 1.5. Beneath material 2 is material 3 with an
index of refraction n3.The ray is incident on the air–material 2 interface at the
Brewster angle for that interface. The ray of light refracted into material 3 happens
to be incident on the material 2–material 3 interface at the Brewster angle for that
interface. What is the value of n3?
150
Lambert’s Law of Absorption
• The amount of light absorbed by a lens is given by Lambert's law of absorption,
which states that for an absorptive material such as a tinted (hafif renkli) lens layers of
equal thickness absorb equal quantities (or percentages) of light regardless of the
intensity of the light.
• The intensity of light through
the lens is calculated as
• where I0 intensity of light
incident upon the first layer of
the lens, Ix total amount of
light lost by absorption and x
is the number of layers
• The total transmission of the
lens, T, will then be
FIGURE: Absorption of light by a lens (Lambert's law).
T=TX- RB
• RB designating the light lost by reflection at the back surface of the lens.
151
Reflection
Example:
Given an ophthalmic lens, 6 mm thick, having a transmission factor, q, of 0.8 per 2
mm of thickness, and an index of refraction of 1.523. What is the transmission, T of
this lens?
The light lost by reflection at the front surface is
The intensity of light incident upon the first layer of the lens, after reflection is
Io = I- RF = 1 - 0.043 = 0.957
The total amount of light lost by absorption will be
I3 = Io(q3) = 0.957(0.83) = 0.957(0.512) = 0.49
The light reflected by the back surface, RB, is
and the total transmission, T, of the lens is equal to
152
Reflection
EXAMPLE:
Taking into account only single reflections, what percent of the incident light is
transmitted by an ophthalmic crown (n = 1.523) lens in air?

4.3% is reflected at each surface. Therefore, 95.7% of the incident light is
transmitted through each surface (see below figure). As a first guess, since the
lens has two surfaces, we would expect the lens to lose 8.6% due to reflection
(i.e., 2 times 4.3%). For a more accurate calculation, let I0, I1? and I2 be the
intensity of the light, respectively, incident on the first surface, transmitted through
the first surface and incident on the second surface, and transmitted through the
second surface (see figure). The transmittance factor at the first surface is 0.957,
or
I1= 0.957I0.
The transmittance factor at the second
surface is again 0.957, or
I2 = 0.957 I1 ;
FIGURE: Reflection losses for a ray passing through a lens.
153
Reflection
EXAMPLE (Cont.):
Thus, the transmittance factor for both surfaces is 0.916, or 91.6% of the light
incident on the lens is transmitted through it. The other 8.4% is lost by reflection.
The 8.4% differs only slightly from the first guess of 8.6%. For interfaces with a
higher percent reflection the first guess is not as close to the answer, whereas for
interfaces with a lower percent reflection the first guess is better.
154
Opacity
• Opacity is defined as the reciprocal of transmission
• If light passes through a number of lenses the ultimate (nihai) transmission is found
by
• and the ultimate opacity is found by
Example:
An automobile driver wears a tinted lens having a transmission of 80%, while looking
through a windshield having a transmission of 70%. What are (a) the ultimate
transmission and (b) the ultimate opacity of the windshield-lens combination?
(a) The ultimate transmission is equal to
TU = (T1)(T2) = (0.80)(0.70) = 0.56 = 56%.
(b) The ultimate opacity is equal to
O = (O1)(O2)
155
Density
• The density may be found by use of the relationship
Density = - log T
• Since the transmission of a medium is the ratio of transmitted light (I) to incident
light (I0), the transmission of each surface can be expressed as a density (D).
• Hence, for a simple lens the total density of the lens may be considered to consist
of the two surface densities and a media density.
• The density of the front surface of the lens is designated by
where I0 = I — R1
• Since the total density of a lens is the sum of the densities of the parts, then the
total density is given by the expression
D = 2 (surface densities) + (number of units)(density per unit)
D = 2 DS + (x)(D/unit)
156
Density
EXAMPLE-1:
A lens 2 mm thick has a refractive index of 1.523 and a total transmission of 30%. If a
transmission of 50% is desired, what must be the thickness of the lens?
The density of the 2-mm-thick lens is found by the formula
D = -log T= -log 0.3 = -(-0.523) = 0.523
The reflection from the front surface is found by
The transmission, T, of the front surface is given by
T = I – R1 = 100 - 4.3 = 95.7%
The front surface density is equal to
Ds = -log T = - log 0.957= -(-0.0191) = 0.0191.
The two surface densities (front and back) are equal to 2(0.0191) = 0.0382.
157
Density
EXAMPLE-1 (Cont.):
The total density of the lens is given by the relationship
0.523 = 2DS + Dmed;
therefore,
Dmed = 0.523 - 0.0382 = 0.4848
Since the lens is 2 mm thick, the density per millimeter of thickness is
0.4848/2=0.2424
The density of a lens with a transmission of 50% is
D = - log T = -log 0.5 = -(-0.301) = 0.301.
D = 2DS + (number of units)(D/unit)
0.301 = 0.0382 + (number of units)(0.2424),
or
(0.2424) (number of units) = 0.301 - 0.0382 = 0.2628.
and,
(number of units) = 0.2628 / 0.2424 = 1.08
The thickness of the lens is therefore equal to 1.08 mm.
158
Density
EXAMPLE-2:
If the above lens is made 3.5 mm thick, what would be its transmission?
Total density = 2DS + (number of units)(D/unit)
= 0.382 + (3.5)(0.2424) = 0.0382 + 0.8484 = 0.8866.
Therefore, the density of this lens, if made in a thickness of 3.5 mm, would be 0.8866.
log T= -D= -0.8866.
The antilog of -0.8866 = 0.1298,
and the transmission of the lens is therefore 12.98%
159
Reflection
• In below figure, five common sources of spectacle lens reflections are shown:
FIGURE: Types of surface reflections:
Note that in A and B the light is incident
upon the back surface of the lens, while
in C, D, and E the light is incident upon
the front surface of the lens.
• The methods available for minimizing the annoyance (can
reflections include:
• base curve selection;
• changing the vertex distance;
• selection of small lens sizes;
• patient education and counseling (rehberlik);
• the use of antireflective coatings.
sıkan)
of surface
160
Rayleigh Scattering
• Several interesting phenomena can occur
when a photon is incident on an atom.
• If the energy of the incoming photon is too
small to excite the atom to an excited state,
the atom remains in its ground state and the
photon is said to be scattered.
• Since the incoming and outgoing, or
scattered, photons have the same energy,
the scattering is said to be elastic.
• If the wavelength of the incident light is
large compared with the size of the atom,
the scattering can be described in terms of
classical electromagnetic theory and is
called Rayleigh scattering after Lord
Rayleigh
• According to Rayleight scattering; the
scattering from atoms and molecules
depend on the wavelength.
161
Rayleigh Scattering
• A detail analysis shows that the
intensity of scattered light depends on
the fourth power of the wavelength:
Iscattering  -4.
• This wavelength dependence explains
why the sky is blue and sunsets are red.
• As sunlight travels through the
atmosphere, the -4 dependence of
Rayleigh scattering causes the shorter
wavelengths
to
be
preferentially
scattered.
• If we take 650 nm as a typical
wavelength for red light and 450 nm for
blue light, the intensity of scattered blue
light relative to scattered red light is
162
Rayleigh Scattering
• A detail analysis shows that the intensity of scattered light depends on the fourth
power of the wavelength: Iscattering  -4.
• This wavelength dependence explains why the sky is blue and sunsets are red.
• As sunlight travels through the atmosphere, the -4 dependence of Rayleigh
scattering causes the shorter wavelengths to be preferentially scattered.
• If we take 650 nm as a typical wavelength for red light and 450 nm for blue light, the
intensity of scattered blue light relative to scattered red light is
• Four time more blue light scattered towards us than red light and thus as shown in
figure, the sky appears blue.
• Because of the earth’s curvature, sunlight has to travel much farther through the
atmosphere when we see it at sunrise or sunset than it does during the midday
hours.
• In fact, the path length through the atmosphere at sunset is so long that essentially
all the short wavelengths have been lost due to Rayleigh scattering.
• Only the longer wavelengths remain-orange and red- and they make the colors of
sunset.
163
ABERRATION
• Virtually all optical systems have aberrations (sapma) that degrade (bozmak) the
quality of the image they create.
• Because of aberrations, the image formed by a optical system is not perfect.
• Aberrations are an important consideration in the design of spectacle lenses
(gözlük camları) and the correction of refractive errors.
• Ophthalmic lenses are generally designed so that certain of the aberrations are
minimized.
164
Aberration
• A yellow-green color (an average wavelength of 587.56 nm), produced by the
helium d (or D3) - line, is the standard reference wavelength for ophthalmic
optics in the United States.
• In some countries, 546.07 nm is used as the reference wavelength, which is
produced by the mercury green-(E)-line.
• This means that the power of a lens or prism is based upon the refractive index
(n) that the material has for the chosen reference wavelength.
• Because the lens’s index of refraction is different for each of the wavelengths
constituting white light, its focal length is different for each.
• The chosen wavelength generally lies close to the peak photopic sensitivity of
the human eye, which is related to the maximum luminous efficiency of the eye.
• The mean dispersion is the difference in refractive indices between the blue and
red ends of the spectrum at specified wavelengths.
165
Types of Aberration
• If an aberration can be produced with a single wavelength of light, it is referred to
as a monochromatic aberration.
• Monochromatic aberation occurs when incident light is not confined to a narrow
bundle (deste) of paraxial (optik eksene yakın) rays.
• Chromatic aberrations occur only with polychromatic light, which is composed of
multiple wavelengths.
• Chromatic aberration is due to the material from which the lens (or prism) is
made, and therefore can be controlled to some extent by the judicious (akla uygun)
selection of materials.
• The monochromatic aberrations, on the other hand, are due less to the properties
of the material than to factors such as
• the size of the lens aperture,
• the angle the incident rays make with the optic axis of the lens,
• and the position of the lens in relation to the eye.
• They are controlled
• by varying the front and back surface powers
• and the thickness of the lens,
• as well as by varying vertex distance.
166
Chromatic Aberration
• In reality, the refractive index of any material varies slightly as a function of the
wavelength (n  ).
• This means that various colors of light will actually have different indices of
refraction in the same lens material.
• This is a result of the fact that colors of light with shorter wavelengths, i.e. like
violet, travel more slowly through most transparent materials than colors with
longer wavelengths, like red.
• The shorter the wavelength of the light, the more it is deviated on refraction.
• Therefore, violet light has a higher index of refraction than red.
nviolet>nblue>ngreen>nyellow>norange>nred
FIGURE: Chromatic dispersion of white light into its component colors by a prism.
167
Chromatic Aberration
• Hence according to the formula m = (n – 1) A, the angle of deviation for the violet
light will be greater than the angle of deviation for the red light.
• Thus, the ray of violet colour bends maximum towards the base of the prism,
while the ray of red colour bends least.
• The angle between the emergent rays of any two colours is called ‘angular
dispersion’ between those colours.
• The angular dispersion between them is


168
Chromatic Aberration
• When white light passes through a thin prism, the ratio of the angular dispersion
between the violet and red emergent rays and the deviation suffered by a mean ray
(ray of yellow colour) is called the ‘dispersive power’ of the material of the prism.
• It is denoted by ω.

• The dispersive power depends only upon the material of the prism, not upon
refracting angle of the prism.
• Greater is its value for a material, larger is the span of the spectrum formed by the
prism made of that material.
• Dispersive power flint-glass is more than that of crown-glass.
169
Chromatic Dispersion
• Chromatic aberration is a result of the inherent (tabiatında
dispersive properties of a lens material.
doğasında olmak)
• When lenses are used in instruments, it is desirable (istemek) to eliminate
chromatic aberration.
• Chromatic aberration is a measure of the color dispersion produced by optical
materials.
• Due to chromatic aberation, a series of coloured images are formed when white
light is incident upon a spherical lens (see below figure).
• This phenomenon is responsible for the chromatic dispersion of white light into
its component colors by prisms and also lenses.
Figure: Chromatic aberration.
170
Duochrome Test
• In clinical practice the chromatic aberration of the eye is made use of in the
duochrome test.
• In the test,
• a myopic eye sees the red letters more clearly than the green (see below
figure)
• while a hypermetropic eye sees the green letters more distinctly (belirgin biçimde).
Figure: Chromatic aberration, emmetropic eye.
P = principal plane, R = retina.
171
Types of Chromatic Aberration
• Chromatic aberration can be considered either as
• longitudinal (axial) aberation
• transverse (lateral) aberration
172
Longitudinal Chromatic Aberration
• The difference in dioptric power for wavelengths of 486 and 656 nm is defined as
the lens’s longitudinal chromatic aberration (CA).
Longitudinal CA = Ff − Fc
• where Ff is the dioptric power for 486 nm, and Fc is the dioptric power for 656 nm.
• Longitudinal chromatic aberration can also be calculated using the Abbe number
(v) and dioptric power for light of 589 nm (Fd) with the following formula:
Figure : Longitudinal chromatic aberration is defined as the difference in dioptric power for wavelengths of
486 and 656 nm.
173
Abbe Number
• To quantify the amount of dispersion produced by a prism or lens, we use three
wavelengths: 486, 589, and 656 nm.
• The refractive indices for these wavelengths are designated as nf, nd, and nc,
respectively.
• When we specify the index of refraction of a substance, we normally specify the
index for the sodium D-line.
• This is the Fraunhofer line having a wavelength of 589 nm.
• The dispersive power (ω ) is given by
• where nD, nF and nC are the refractive indices of the material at the wavelengths of
the Fraunhofer D-, F- and C- spectral lines (589.3 nm, 486.1 nm and 656.3 nm
respectively).
• The dispersive power of BK7 glass is 0.015595.
• As this value increases, the dispersion of the prism or lens increases.
174
Abbe Number
• The reciprocal of the dispersive power is defined as the refractive efficiency, or
constringence (also called the Abbe number or v) of a transparent refracting
material:
• v is the Abbe number for d, F and C wavelengths.
• These are the usual wavelengths used for correcting the dispersion or chromatic
aberration of visual optical systems.
• Other v numbers are used when correcting chromatic aberration in other parts of
the visible, infrared or ultraviolet spectrum.
• BK7 glass has an Abbe number v (or Vd) of
• The quantity nF - nc is the mean dispersion.
• It is 0.00806 for BK7 glass.
175
Abbe Number
• In comparing the nu (v) values of two media,
• if the media have the same mean dispersion; the one having the higher
mean refractivity will have the higher nu value,
• if the two media have the same refractivity; the medium having the lower
mean dispersion will power.
• Abbe numbers are used to classify glass and other optically transparent
materials.
• The most desirable lens materials are generally those with the higher nu (v)
values.
• In the visible spectrum, Abbe numbers of optical glasses range from 20 to 100.
• The glasses are divided into two broad categories, known as crowns and flints.
• For example, the nu (v) value of ophthalmic crown glass is 59, whereas that of
flint glass is 30.
• Typical values of v range from around 20 for very dense flint glass, around 30 for
polycarbonate plastics, and up to 65 for very light crown glass, and up to 85 for
fluor-crown glass.
• An Abbe refractometer is a bench-top device (tezgah
precision measurement of an index of refraction.
üstü cihaz)
for the high-
176
Abbe Number
• Below figure illustrates the variation in refractive indexes with the wavelength of
BK7 (crown), F4 (flint) and SF8 (dense flint) glasses.
• The flint glasses have greater indices than the crown glass.
• Further, the flint glasses are more dispersive, i.e., the rate of change in their
slopes increases more rapidly compared with the crown glasses, especially, at
the short wavelengths.
Figure: Variation in refractive index with glass type.
177
Abbe Number
• A six-digit code number is used to designate optical glasses.
• For example the number 517642 identifies BK7 glass.
• The first three digits, 517, are the rounded off first three digits following the
decimal point in nd.
• The last three digits, 642, are the first three digits of the Abbe number obtained
by omitting the decimal point.
• A flint glass, identified as F4 has a code number of 617366.
• Its index nd = 1.61659 and Vd (or v) = 36.63.
178
Abbe Number
• Optical glass manufacturers plot the variety of glasses they make on a nd/Vd
diagram or glass map such as in below figure.
Figure: A glass map showing the indices and dispersions of optical glasses manufactured by the Schott
Optical Glass Co.
179
Abbe Number
• Crown glasses largely have lower refractive indices and dispersive power than
flints.
• Crown glasses have nd >1.60, and Vd >50, or nd<1.60 and Vd>55.
• Some optical data for BK7 and flint glass - F4 are shown in below table.
• The BK7 glass belongs to the borosilicate crown group.
• BK7 is the abbreviation used by the Schott Optical Glass Co. To identify
borosilicate crown glass No 7.
Table: Indices of refraction of a crown and a flint glass.
180
Longitudinal Chromatic Aberration
• The formula for the longitudinal chromatic aberration of a lens can be derived by
using the lens maker's formula for the Fraunhofer D, F, and C-lines:
D  589.3 nm
F  486.1 nm


C  656.3 nm

• Where fC and fF are the red and blue focal lengths; uc‘and uF' are the red and blue
image distances
• This expression shows that the Iongitudinal chromatic aberration of a lens varies
directly with power.
• For example, for a —6.00 D lens having a nu (v) value of 60
181
Longitudinal Chromatic Aberration
EXAMPLE: A thin equiconvex lens with radii of 100 mm is made of BK7 glass. It
has a focal length fd = 96.749 mm and its power is +10.336 D. Find the longitudinal
chromatic aberration in diopters.
From below equation
we find, LCA = 10.336/64.2 = 0.161 D.
EXAMPLE: We can find the linear LCA of the thin BK7 equiconvex lens by finding
the focal lengths fd= 96.749 mm, fF = 95.716 mm, and fc = 97.215 mm and solving
for
LCA = fc - fF = 97.215 - 95.716 = 1.499 mm.
Alternatively,
LCA = f/V = 96.749/64.2 = 1.507 mm.
182
Longitudinal Chromatic Aberration
EXAMPLE: If the object for the BK7 lens (fd= 96.749 mm) is two focal lengths
away, i.e., ud = -2(96.949) = -193.498 mm, we can find the longitudinal chromatic
aberration by solving the thin lens equation for uC' with the red focal length and uF'
with the blue focal length.
183
Transverse (Lateral) Chromatic Aberration
• The chromatic aberration produced by a prism is referred to as lateral
chromatic aberration (CA), or transverse chromatic aberration (TCA), or
chromatic power.
• We can define it as the difference in prismatic power for wavelengths of 486 and
656 nm.
Lateral CA = Pf − Pc 
• where Pf is the prismatic
power for 486 nm, Pc is the
prismatic power for 656 nm
and Pd is the prismatic
power for 589 nm.
Figure : As the lens’s prismatic power increases toward the periphery of the lens, the lateral chromatic
aberration also increases.
184
Transverse (Lateral) Chromatic Aberration
• For lenses, the transverse chromatic aberration of a lens can also be
expressed in terms of the differences in prismatic effects created by blue and red
light.
• The transverse chromatic aberration of a lens may also be expressed as follows:

FIGURE: Transverse CA expressed as the difference in prismatic effects created by red and blue light.
185
Transverse (Lateral) Chromatic Aberration
• Since the secondary focal length of a lens will be different for each of the
monochromatic constituents (bileşen) of white light,
• the axial image positions of these constituents will differ
• and hence, for an extended object, their image sizes will differ (see below figure).
• Terefore, the transverse chromatic aberration is often referred to as the
chromatic difference in magnification.
FIGURE: Transverse CA is also referred to as the chromatic difference in magnification.
186
Transverse (Lateral) Chromatic Aberration
• Transverse chromatic aberration in prism diopters is equal to the difference in the
blue and red deviations.

EXAMPLE:
A white light ray from a star at 5.711° strikes the thin lens of the previous example
at a height h = 20 mm. Find the Transverse chromatic aberration in prism diopters.
Note: V = 64.2, f = 96.749, thus, F= 10.336 D.

On looking through the lens at a height of 20 mm the image of the star will be a
spectrum that subtends 0.322Δ, or about 11 minutes of arc.
187
Transverse (Lateral) Chromatic Aberration
EXAMPLE:
Find the TCA produced by the thin (96.749 mm focal length) lens, if an infinitely
distant object subtends 5.7106°.

EXAMPLE:
A positive equiconvex lens, made of BK7 glass, has radii of 100 mm and a
thickness of 20 mm. Find the Transverse chromatic aberration in mm for a star at
an angle of 5.7106° to the optical axis.
The equivalent focal length of the lens is 100.1619 mm. The V number is 64.167.
According to equation:

188
Transverse (Lateral) Chromatic Aberration
• Transverse chromatic aberration is the more important chromatic aberration to
correct in oculars, because they cover wider fields than objectives.
• For example, a 50X telescope with a 1° real field of view requires an ocular with a
50° apparent field of view.
• If the ocular is not corrected, color fringes surrounding the image of an extended
object will be visible.
• Because transverse chromatic aberration is a chromatic variation in magnification
due to a chromatic variation in focal length, it can only be corrected by a lens
system with a constant focal length or power in all wavelengths.
189
Angular Dispersion
• Since a prism does not change the vergence of light, it does not manifest (göstermek)
longitudial (axial) chromatic aberration.
• It does, however, manifest angular dispersion (similar to lateral chromatic
aberration).
• The angular dispersion is defined as the difference in prismatic effects for blue
and red light (Fraunhofer F, C and D), that is,

• This equation tells us that the chromatic (angular) dispersion of a prism is equal
to the deviation of the D-line divided by the nu (v) value of the prism.
FIGURE: A prism can manifest transverse CA but cannot manifest longitudinal chromatic aberration.
190
Angular Dispersion
• The dispersion of d, F and C light by a thin prism of refracting angle ßΔ is shown
in below figure.
• The deviation of these rays in prism diopters is PdΔ, PFΔ and PCΔ .
• Where,
Figure: Chromatic aberration of a prism in prism diopters.
• The chromatic aberration of the prism is the deviation in blue minus the deviation
in red light.
191
Angular Dispersion
• Thus, the chromatic aberration of the prism is equal to the mean dispersion times
the refracting angle.
• According to below equation



• A prism of P = 10Δ, made of BK7 glass (517642), will have a chromatic aberration
of 10/64.2 = 0.156Δ.
• The angle between the red and blue light will be 0.156A or 5.35 minutes of arc.
• This chromatic blur is about the size of a 20/20 letter.
• A 10Δ prism made of F4 glass (617366) will produce a blur of 9.39 minutes of arc.
192
Paraxial Zone
• The prismatic effect of a spherical lens is least in the paraxial zone and increases
towards the periphery (kenar) of the lens.
• Thus, rays passing through the periphery of the lens are deviated more than those
passing through the paraxial zone of the lens (see below figure).
• Gaussian Optics is rays strike the lens about a region close to the optical axis so
that all rays form a perfect image without aberrations.
• This may also be referred to as first order optics because the angles the ray
makes with the axis are small and the sine of the angle equals the angle in
radians.
• Higher order optics (3rd, 5th, 7th, etc.) have aberrations.
193
Prismatic Effect of Thin Lenses
• The amount of bending an incident ray undergoes after refraction through a thin
lens is a function of the distance of the ray from the optical axis.
• If an incident ray close to the axis bends less than a peripheral incident ray
because both are refracted toward the same secondary focal point.
• This change in the direction of the incident ray is the deviation, and it may be
considered to be the prismatic effect of the lens.
• Ass seen in below figure, prisms have one deviation for any position of the incident
ray 1 = 2.
FIGURE: Prismatic effect of a thin lens and a prism. The deviation increases toward the periphery of
the thin lens. The deviation of a prism is constant no matter where on the prism the ray is incident.
194
Prismatic Effect of Thin Lenses
• Unlike prisms, a thin lens may be considered to be an infinite series of prisms with
increasing deviation (and apical angles) toward the periphery (1 > 2).
• The direction of the prismatic effect may be described in terms of the base.
• There is no prismatic effect at the optical center of the lens.
• However, if a lens is decentered (merkezden ayrılmak), a prismatic effect is induced.
• If the positive lens is decentered downward so that one views through the top
portion of the lens.
• The effect will be similar to a base-down prism, and the image will appear to move
up (see below figure).
FIGURE 2: Image seen through a prism.
FIGURE: Location of image formed through decentered lens.
195
Prismatic Effect of Thin Lenses
• If the negative lens is decentered downward so that one views through the
upper portion of the lens.
• The effect will be similar to a base-up prism, and the object will appear to move
down (see below figure).
• Positive lenses will displace images opposite the direction of decentration, and
negative lenses will displace images in the same direction as decentration.
FIGURE: Location of image formed through decentered lens. Notice the base of the prism.
196
Achromatic Prisms
• The prismatic effect is used in hand neutralization of lenses where the movement
due to the prismatic effect is neutralized by another lens of the same power
magnitude but opposite sign.
• An achromatic prism is a prism that produces deviation without dispersion, and is
constructed by placing two prisms made of different materials in direct contact with
the bases oriented in opposite directions.
• Such a prism can be made achromatic for two wavelengths, usually for the F and
C-lines.
• The rays of light for the two wavelengths for which the prism is achromatic will
emerge parallel, as shown in below figure.
FIGURE: An achromatic prism.
197
Achromatic Prisms
• The condition for achromatism is that the chromatic aberrations of the two prisms
are equal.
• The combination of a crown prism P1 with a weaker flint prism P2 will have a net
power P
P = P1 + P2
• Because their bases are opposite, the aberrations will cancel.
• This can be stated as follows:


• The deviating powers of the two prisms are given by the derived expressions
and
which can be used to design an achromatic prism.
198
Achromatic Prisms
EXAMPLE: Design a 10Δ achromatic prism with BK7 and F4 glasses. See below
figure.
The glass data are
BK7 (517642) nd = 1.51680, Vd = 64.2
F4 (617366) nd = 1.61659, Vd = 36.6
Figure: An achromatic prism.


The negative sign means that the base of the flint prism is opposite that of the
crown.
199
Achromatic Prisms
EXAMPLE (Cont.):
The refracting angles of the prisms
The refracting angles have been computed for index nd. To check on the
achromatism provided by these prisms, calculate the deviation in F and C light with
the following equations:
Deviation in F light:
Deviation in C light:
The achromatic prism deviates red and blue light identically (9.9951Δ) but different
from yellow light (10Δ). This slight error of 0.005Δ, due to the irrationality of
dispersion, is called secondary color.
200
Achromatic Prisms
Example:
Given two prisms, one having an index of refraction of 1.50 and a nu (v) value of 60,
and the other having an index of refraction of 1.60 and a nu (v) value of 40. Design
an achromatic prism having a power of = 2 prism diopters.

and

201
Achromatic Lenses
• In order for a lens to be achromatic, two lenses of different
materials, one convex and one concave, having different nu (v)
values, must be used (see below figure).
• The power of each lens must be such that
Fig: An achromatic lens.
202
Achromatic Lenses
Example:
Given two lenses, one having an index of refraction of 1.50 and a nu value of 60,
and the other having an index of refraction of 1.60 and a nu having an index of
refraction of 1.60 and a nu value of 40. Design an achromatic lens having a
refracting power of -2.00 D.
203
Achromatic Lenses
EXAMPLE: Design an achromatic doublet with a power of 10 Diopters. Use BK7
and F4 glass. Given: F=10 D, V1 = 64.1669, and V2 = 36.6146. The power of the
crown lens from below equation:
 F1 = +23.2891 D.
Therefore, F2 = -13.2891.
Let the lenses be piano convex and piano concave. The radii of the convex r1 and
concave r3 surfaces are
The powers of the crown and flint lenses in blue light are
204
Achromatic Lenses
EXAMPLE (Cont.):
The power of the doublet in blue light is
For red light Fc = +9.9951 D.
The red and blue light comes to a
common focal plane at 100.0489
mm from the lens. Yellow is focused
at 100 mm (see figure). All the
intermediate wavelengths focus
between these extremes. The slight
difference in focus is called the
secondary spectrum. Figure shows
the
longitudinal
chromatic
aberrations of the BK7 singlet and
the achromatic doublet.
Figure: Longitudinal chromatic aberration of a
singlet and an achromatic doublet.
205
Problems
1. Find the angles of deviation in d, F and C wavelengths by a 15Δ prism, made of
BK7 glass, and the chromatic aberration of the prism in minutes of arc. Ans.: PdΔ
= 15Δ, PFΔ = 15.162Δ, PCΔ = 14.928Δ, Chr. Ab. = 0.234Δ = 8.04'.
2. Find the chromatic aberration (in
glass 517642. Ans.: 0.234Δ, 8.04‘
Δ
and minutes of arc) of a 15Δ prism made of
3. Find the angles of deviation in d, F and C wavelengths by a 9Δ prism, made of F4
glass, and the chromatic aberration of the prism in minutes of arc. Ans.: PdΔ =
9Δ, PFΔ = 9.174Δ, PCΔ = 8.928Δ, Chr. Ab. = 0.246Δ = 8.45'.
4. Find the chromatic aberration (in
glass 617366. Ans.: 0.246Δ, 8.45‘
Δ
and minutes of arc) of a 9Δ prism made of
5. Design an achromatic prism with a power of 10Δ using glasses: crown (500600)
and flint (600200). Ans.: crown = 15Δ; flint = 5Δ.
6. An achromatic prism is made with a 16Δ prism made of glass 550500 and an 8Δ
prism made of flint glass with a mean dispersion of 0.026. Find the power of the
achromat; and the index of refraction and V number of the flint glass? Ans.: 8Δ;
1.650; 25.
206
Problems
1. 8. A thin lens has a focal length in d wavelength of 100 mm. The lens indices in
d, F and C wavelengths are: 1.51, 1.52 and 1.505. Find its focal length in F and
C wavelengths, and LCA. Ans.: fF =98.077 mm; fc=100.990 mm; LCA=+2.913
mm.
2. A 100 mm focal length lens is made of glass with a V number of 34. Find LCA
for an object at infinity. What is the V number of the glass in Problem 8? Ans.:
2.941 mm; 34.
3. 10. An object is located two focal lengths in front of the lens in Prob. 8. Find the
image distances in d, F and C wavelengths and the LCA in the image. Ans.: u'd
= 200mm; u'F = 192.4528 mm; u'c = 204 mm; LCA=+11.547 mm.
207
Types of Monochromatic Aberrations
• Lens design generally addresses the monochromatic aberrations which are
independent of color.
• There are five unique monochromatic aberrations.
• They are:
• Coma,
• Spherical Aberration (Longitudinal and Transverse),
• Oblique Astigmatism,
• Curvature of the field,
• Distortion.
208
Spherical Aberration
• Spherical aberration affects object points along the optical axis (see below figure).
• Rays of light refracted in the periphery of a lens come to a different focal point
than central rays—producing spherical aberration.
• The paraxial approximation for lens power, which assumes a focus at point F',
loses accuracy for these peripheral rays.
FIGURE: Spherical aberration.
209
Longitudinal Spherical Aberration
• Longitudinal spherical aberration is minimized for lenses with approximately
planoconvex shapes that are oriented so that the front surface is more convex.
Figure: For a planoconvex lens, longitudinal spherical aberration is minimized when light rays are incident on
the convex surface of the lens, but not on the plano surface.
210
Longitudinal Spherical Aberration
• The amount of longitudinal spherical aberration is especially dependent on the radii
of curvatures of the front and back surfaces of a lens.
• These can be used to calculate the Coddington shape factor, σ, which is related
to the amount of spherical aberration
• where r1 is the front surface’s
radius of curvature and r2 is the
back (ocular) surface’s radius of
curvature.
• Figure shows longitudinal spherical
aberration plotted as a function of
the shape factor for different shape
lenses all of which have the same
power.
Figure : Longitudinal spherical aberration as a
function of the shape factor.
211
Lateral Spherical Aberration
• Lateral (or transverse) spherical aberration is specified in terms of the diameter of
the confusion disc at the paraxial focus.
• The confusion disc being defined as the image formed on a screen placed at any
point along the caustic.
212
Coma
• Coma is similar to spherical aberration but affects object points off the optical the
axis.
• Spherical aberration occurs for beams of light that are parallel to the optic axis of a
lens.
• However, coma occurs for oblique (eğik) beams (see below figure).
• Coma occurs as a result of the fact that each zone of a lens has its own focal
length.
FIGURE: Coma.
213
Reducing of Spherical Aberration
• Spherical aberration is a problem only for large aperture (diafram) optical systems:
• Longitudinal spherical aberration increases with the square of the aperture.
• Lateral spherical aberration increases with the cube of the aperture.
• Therefore, the size of the aperture should be minimized to reduce spherical
aberration.
• In addition to controlling the size of the aperture, another technique of reducing
spherical aberration is to employ a doublet (çift mercek kullanmak).
214
Reducing of Spherical Aberration
• If achromatic lens systems are composed of lenses of varying material combined
so that the dispersion is neutralised while the overall refractive power is
preserved.
• For example, by combining a convex lens of high refractive power and low
dispersive power with a concave lens of low refractive power but higher
dispersive power, the aberration can be neutralised while preserving most of the
convex lens refractive power.
Figure: Diagram showing the principle of the aspheric doublet lens.
215
Reducing of Spherical Aberration
• Lens form may also be adjusted to reduce spherical aberration, e.g. plano-convex
is better than biconvex.
• For example: spherical aberration may be controlled by varying the "bend" of the
lens, that is, by manipulating the curvature of each of the two surfaces without
changing the refracting power of the lens.
• To achieve the best results, spherical surfaces must be abandoned (kullanmayı
bırakmak) and the lenses ground with aplanatic surfaces, that is, the peripheral
curvature is less than the central curvature.
Figure: Aplanatic (aspheric) curve to correct spherical aberration.
216
Reducing of Coma and Spherical Aberration
• Spherical aberration is not of great importance for ophthalmic lenses of moderate
power, since the eye uses only a small portion of the lens at any one time due to
the limiting aperture of the pupil.
• However, for lenses of high power (+10.00 D or more), it can be serious.
• As a practical matter, spherical aberration is taken into consideration in ophthalmic
lens design only for high-powered convex lenses, for which plastic aspheric lenses
are routinely used.
• Like spherical aberration, coma is a problem mainly for large-aperture optical
systems.
• Therefore, Coma can be largely ignored in the design of spectacle lenses because
of the limiting effect of the pupil.
• In addition to aperture size, the following factors can be used to control coma.
• They are the form of the lens and the angle of obliquity (eğim, meyil).
217
Distortion
• Distortion is an aberration that affects not the focal quality of an image, but its
size and shape (or its geometric reproduction).
• If there is no distortion present in the image, the image is said to be orthoscopic,
(as illustrated in below figure).
• For plus-powered lenses, excess magnification in the periphery of the lens causes
pincushion distortion (içbükey bozulma).
• For minus-powered lenses, excess minification causes barrel distortion (fıçı
bükülmesi).
• Distortion causes objects to appear misshapen (bozuk
especially in higher lens powers.
şekilli)
and curved—
FIGURE: Distortion. Orthoscopy is a lack of distortion. Minus lenses produce barrel distortion. Plus lenses
produce pincushion distortion.
218
Optical Aberration
• Like refractive errors, optical aberrations cause imperfect image formation and
blurred vision (bulanık görme).
• For ophthalmic lenses, an optical aberration occurs when rays of light fail to
come a point focus at the intended position of the far point MR of the eye as it
revolves about its center of rotation at point R.
FIGURE-1: Lens design parameters for a plus lens. The far point MR is rotated about the center of rotation R,
creating a spherical surface called the far-point sphere. The radius of the far-point sphere rFPS = h - f', where h
is the stop distance from the lens to the center of rotation.
219
Aberration
• When viewing peripherally (kenarından) through a lens, we want the focus to fall on
the far point sphere.
• However, it should be apparent that the base curve, or “bending” of the lens, will
affect its distance from the eye and its tilt (eğilmek) relative to the line of sight
(görüş).
• These factors can significantly alter the power that the eye encounters, resulting
in a power that deviates from that required for clear vision.
FIGURE: Lens design parameters for a minus lens. The same descriptors from plus lenses apply.
220
Oblique Astigmatism
• Oblique astigmatism and power error occur as the spectacle-wearer gazes (Gözü
belli bir noktadan ayırmaksızın bakmak) away from the optical axis—or optical center—of
the lens.
• Consequently, peripheral (merkezden dışarı doğru) vision through a lens that suffers
from an excess of either of these two aberrations is blurred (bulanık), and the wearer
experiences a limited field of clear vision.
• This effect is illustrated in below figures for a 30° angle of view from the optical axis
of the lens.
FIGURE: Clear peripheral vision at a 30° viewing angle
with a steeper lens form.
FIGURE: Blurred peripheral vision at a 30° viewing
angle with a flatter lens form.
221
Oblique Astigmatism
• Astigmatic focusing error results when rays of light (ışık hüzmesi) from an off-axis object
strike the lens obliquely.
• Two focal lines are produced from a single object point.
• The dioptric difference between these two foci, FT - FS, is the amount of the oblique
astigmatism, or the oblique astigmatic error A.
A = F T - FS
• Oblique astigmatism is also referred to as marginal or radial astigmatism.
• Oblique astigmatism is one of the principal lens aberrations that must be corrected
for when designing spectacle lenses.
FIGURE: Oblique astigmatism. Rays of light from an object point strike the lens obliquely and are focused into
two separate line foci, instead of a single point focus, when oblique astigmatism is present.
222
Oblique Astigmatism
• Rays of light striking the tangential (or meridional) plane of the lens come to a line
focus at the tangential focus FT.
• The resultant focal line is perpendicular to the actual tangential plane.
• Rays striking the sagittal (or equatorial) plane come to a line focus at the sagittal
focus FS.
• This focal line is perpendicular to the sagittal plane (Both
previous figure).
of these planes are shown in
• The actual tangential power error PT is equal to the difference between the desired
back vertex power FV and the tangential power FT,
PT = FV - FT.
• Similarly, the sagittal power error PS is equal to the difference between the desired
back vertex power the sagittal power FS,
PS = FV - FS.
• Therefore, the astigmatic error A is also given by,
A = PT - PS = FT - FS.
223
Questions
1. An object is 24 cm in front of a plane refracting surface separating water (n =
4/3) from glass (n' =1.5). How far from the surface will a ray with a slope of 60°
cross the axis, after refraction? Repeat for slopes of 45°, 30° and 0.001°. Draw
a ray diagram to scale. Ans.: -34.467 cm; -29.928 cm; -27.928 cm; -27.000 cm.
2. A ray makes an angle of incidence of 30° on a prism of index 1.75, with a 45°
refracting angle: Find the angles of: (a) refraction at the first surface, (b)
incidence and (c) emergence at the second surface, (d) deviation; (e) critical
angle; f) minimum angle of incidence; g) angle of incidence for minimum
deviation; and h) the minimum deviation angle. Construct the paths of the 30°
ray and the minimum deviation ray. Ans.: (a) 16.602°; (b) -28.398°; (c) -56.336°;
(d) 41.336°; e, 34.850°; (f) 17.963°; (g) 42.044°; (h) 39.087°.
3. A ray, normally incident on a prism of index 1.5, has an angle of emergence of48.590°. Find: (a) the prism angle, and (b) the deviation. Construct the path of
the ray. Ans.: (a) 30°; (b) 18.590°
4. The index of the prism in previosu problem 3, for red light is 1.49, and for blue
light 1.52. Find the angle between the red and blue light (this is called the
dispersion). Ans.: 1.305°
224
Questions
1. A ray through a 30° prism undergoes a minimum deviation of 32.348°. (a) What
is the index of the prism? (b) the angle of incidence (c) the limiting angle of
incidence? Ans.: (a) 2; (b) 31.174°; (c) 0°
2. What must be the angle of incidence on a 45° prism of index 1.6, to obtain
grazing emergence? Ans.: 10.14°
3. What is the minimum index of refraction of a 60° prism so that no rays will be
transmitted?
4. What is the minimum refracting angle of a prism of index 1.75 so that no rays
will be transmitted? Ans.: 69.7°
5. White light makes an angle of incidence of 30° at the first face of a 60° prism. Its
index is 1.500 for red light and 1.505 for blue light. Find the angular dispersion.
Ans.: 1.2364°
6. Rays traveling in air strike the first face of a 45° prism of index 1.5 with an
incident angle of 30°. The rays emerge into air. Find: (a) the angle of deviation
at the first surface; (b) the angle of incidence at the second surface; (c) the
angle of emergence and (d) the total deviation. Construct ray path. Ans.: (a) α1 =
10.529°; (b) α2 =-25.529°; (c) α'2 = -40.274°; (d)  = 25.274°
7. Rays travelling in air strike the first face of a 60° prism of index 1.5 with an
incident angle of 45°. The rays emerge into air. Find: (a) the angle of emergence
and (b) the total deviation. Construct ray path. Ans.: (a) -52.381°; (b)  = 37.381°
225
Questions
1. Where are the focal points of a plane spherical refracting surface and a plane
mirror?
2. A glass rod of index 1.5 has a convex radius of curvature of 50 mm on its front
end. Its rear end is ground flat, but not polished. An image of the sun is formed
on this end. How long is the rod? Ans.: 150 mm.
3. The above rod is placed in a swimming pool. Find: a) the object distance that is
conjugate to the ground end. b) the lateral magnification. Ans.: a) virtual object
at 200 mm, b) 0.667.
4. An object 10 cm high, is 100 cm in front of a concave surface of 50-cm radius
that separates air from glass of index 1.75. Find a) the image distance, b) the
size of the image. Construct image Q'. Ans.: u' = -70 cm, y' = +4 cm.
5. A glass ball of index 1.65 is 250 mm in diameter. How far is the image of the
front vertex of the ball from the second vertex? Ans.: u' = -714.29 mm
6. Find the first and second focal lengths of a spherical refracting surface of radius
+75 mm, separating air from index 1.85. Ans.: f= 88.235 mm, f= -163.235mm
7. An object is 100 mm in front of a spherical refracting surface separating air from
glass of index 2. A real inverted image is formed 200 mm from the surface. Find
the radius of curvature. Ans.: +50 mm.
226
Questions
1. A virtual object is located 200 mm from a spherical refracting surface of+50 mm
radius. A real image is formed at 80 mm. The surface separates air from glass.
Find the index of the glass. Ans.: 2
2. A virtual object 7.5 mm high is 75 mm from a concave spherical refracting surface
of radius 25 mm. The surface separates air from glass of index 1.5. Find the
position and size of the image. Construct Q'. Ans.: Virtual, inverted image at u' = 225 mm, y' = - 15 mm
3. A simplified optical model of the eye (a reduced eye) treats the cornea as a
spherical refracting surface separating air from aqueous humor of index 1.336.
Let the radius of this surface be 8mm. a) How long is the eyeball if the retina is
conjugate with infinity (emmetropic)? Find: b) the focal length, and c) the power of
the eye. Ans.: a) 31.81 mm, b) 23.81 mm, c) 42 D.
4. A reduced eye is 24 mm long. The radius of the cornea is 7 mm. a) How far is the
object plane that is imaged on the retina (the far point)? b) Is the eye myopic or
hyperopic? c) How many diopters is the error? Ans.: a) 130.435 mm, b)
hyperopic, c) 7.67 D.
5. When you look into the eye in Prob. 11, you see the pupil of the eye. It appears to
be 3mm from the cornea, and appears 3 mm in diameter, a) How far is the iris
from the cornea? b) How large is its aperture? Ans.: a) 3.503 mm, b) 2.622 mm.
227
Questions
1. A fish is at the center of a spherical fish bowl, 12 inches in diameter, filled with
water. You observe the fish from a distance of 10 inches from the vertex of the
bowl. The fish appears to subtend 5°. How long is the fish? Ans.: 1.05 in.
2. A spherical refracting surface separating air from glass of index 1.6, has a focal
length of 100 mm. Find the position of an image of an object that is 200 mm in
front of the surface. Ans.: 320 mm.
3. When water is the first medium in the above problem, the focal length becomes
300 mm. Find the image distance for the 200 mm distant object. Ans.: -720 mm
4. A +64 mm radius spherical refracting surface separates air and glass. Where
must an object be placed to obtain an inverted image of the same size as the
object? Ans.: 256 mm in front of the surface.
5. A distance of+420 cm separates a real object and its real image, produced by a
spherical refracting surface of radius 35 cm. The surface separates air from
glass of index 1.5. How far is the object? Ans.: u = -105 cm or - 280 cm.
6. A 5-mm high object is 45 mm in front of a concave mirror of radius 20 mm. Find
the position and size of the image. Construct Q'. Ans.: u' = -12.857 mm, y' = 1.429 mm. The image is real and inverted.
228
Questions
1. A spherical mirror projects an image of a flashlight bulb onto a wall 200 inches
away. The bulb is at a distance of 0.667r from the mirror, and the image is 3
times as large as the object. Find the radius of curvature of the mirror. Ans.: -100
in.
2. A real object is midway between the center of curvature and focal point of a
concave mirror. What is the lateral magnification of the image? Construct Q'.
Ans.: -2
3. A virtual object is midway between the center of curvature and focal point of a
convex mirror. What is the lateral magnification of the image? Construct Q'. Ans.:
-2
4. A real object is midway between the focal point and vertex of a concave mirror.
What is the lateral magnification of the image? Construct Q'. Ans.: +2
5. How large is the image produced by reflection from the cornea of a 6-ft high
window, if the window is 12 ft away, and the radius of the cornea is 7.25 mm?
Ans.: 1.80 mm.
6. A light bulb is 5 meters from a projection screen. What is the radius of curvature
of a mirror that will project an image of the bulb magnified 6 times? Ans.: 171.43
cm. concave.
229
Questions
1. A 4-inch diameter convex rear-view mirror has a radius of 12 inches and is 24
inches from a driver. Find the real and apparent fields of view. Ans.: 45.24°,
9.53°.
2. An object 10 cm high, is 100 cm in front of a concave surface of 50 cm radius
that separates air from glass of index 1.75. Find a) the image vergence, b) the
image distance. Ans.: -2.50 D,-70.cm.
3. Find the refracting power of a spherical refracting surface of radius +75 mm,
separating air from index 1.85. Ans.: F= 11.33 D
4. An object is 100 mm in front of a spherical refracting surface separating air from
glass of index 2. A real inverted image is formed 200 mm from the surface. Find
the refractive power of the surface. Ans.: 20 D.
5. A 7.5 mm high virtual object is 75 mm from a concave spherical refracting
surface of 20 D power. The surface separates air from glass of index 1.5. Find
a) the vergence, b) the position, and c) the size of the image. Ans.: a) -6.67 D,
b) Virtual, inverted image at u' = -225 mm, c) y' = - 15 mm
6. A 5-mm high object has a vergence of-22.222 D at a concave mirror of 100 D
reflective power. Find the position and size of the image. Ans.: u' = -12.857 mm,
y' = -1.429 mm.
230
Questions
1. A plane wavefront strikes a +10 D refracting surface separating air from index 1.5.
What is the curvature of the wavefront immediately on entering the image space
medium? Ans.: 6.67 m-1.
2. Two spherical refracting surfaces on an optical axis are separated by a distance of
30 mm. The surfaces form a thick lens containing glass of index 1.5 between the
surfaces. Air surrounds the lens. With respect to the incident light, the first surface
is convex. It has a radius of curvature of 100 mm. The second surface is concave
and has a radius of 200 mm. A real object, 10 mm high, is 400 mm from the front
surface. Find: a) the image distance u2' = A2M2', b) the size of the final image, c)
the refracting power of surfaces 1 and 2. Ans.: a) +194.87 mm, b) -5.13 mm, c)
+5.00 D and + 2.50 D.
3. Two spherical refracting surfaces on an optical axis are separated by a distance of
30 mm. The surfaces form a thick lens containing glass of index 1.5 between the
surfaces. Air surrounds the lens. With respect to the incident light, the first surface
is concave. It has a radius of curvature of 100 mm. The second surface is convex
and has a radius of 200 mm. A 10 mm high real object is 400 mm from the front
surface. Find: a) the image distance u2' = A2M2', b) the size of the final image, c)
the refracting power of surfaces 1 and 2. Ans.: a) -110.84 mm, b) +2.41 mm, c) 5.00 D and -2.50 D.
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Questions
1. Two spherical refracting surfaces on an optical axis are separated by a distance
of 30 mm. The surfaces form a thick lens containing glass of index 1.5 between
the surfaces. Air surrounds the lens. With respect to the incident light, the first
surface is convex. It has a radius of curvature of 100 mm. The second surface is
convex and has a radius of 200 mm. A 10 mm high real object is 400 mm from
the front surface. Find: a) the image distance u2' = A2M2', b) the size of the final
image, c) the refracting power of surfaces 1 and 2. Ans.: a) +7,600.00 mm, b) 200.00 mm, c) +5.00 D and -2.50 D.
2. Two spherical refracting surfaces on an optical axis are separated by a distance
of 30 mm. The surfaces form a thick lens containing glass of index 1.5 between
the surfaces. Air is in front of the lens; water is behind the lens. The first surface
is convex. It has a radius of curvature of 100 mm. The second surface is concave
and has a radius of 200 mm. A real object, 10 mm high, is 400 mm from the front
surface. Find: a) the image distance u2' = A2M2', b) the size of the final image, c)
the refracting power of surfaces 1 and 2. Ans.: a) +384.81 mm, b) -7.59 mm, c)
+5.00 D and + 0. 833 D.
3. Two spherical mirrors are 50 mm apart. An object, 3 mm high, is 20 mm in front of
one mirror. This mirror is convex and has a 10-mm focal length. After reflection
the rays strike the other mirror. It is concave and has a focal length of 10 mm.
Find the image distance and size after reflection by both mirrors. Construct image
point Q2. Ans.: the inverted image is 12.143mm to the right of the second mirror
and 0.2143 mm high.
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TEŞEKKÜRLER
Prof.Dr.A.Necmeddin YAZICI
University Of Gaziantep,
Optic and Acoustic Engineering
mail:[email protected]
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