Download Chapter 19 WAVE PROPERTIES OF LIGHT

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Chapter 19
WAVE PROPERTIES OF LIGHT
GOALS
When you have mastered the contents of this chapter, you will be able to achieve the
following goals:
Definitions
Define each of the following terms, and use it in an operational definition:
interference
coherent source
optical path length
polarization
diffraction
optical activity
noncoherent source
birefringence
dispersion
Application of Wave Properties
Explain the physical basis for: thin film colors, polarimetry, diffraction grating
spectrometry, holography and resolving power of optical systems
Problems Involving Wave Properties
Solve problems involving interference, diffraction, and polarization.
Optical Activity
Design an experimental system capable of measuring the optical activity of a solution.
Lasers
Compare the laser with other light sources in terms of their optical characteristics.
PREREQUISITES
Before you begin this chapter you should have achieved the goals of Chapter 16,
Traveling Waves, and Chapter 18, Optical Elements.
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Chapter 19
WAVE PROPERTIES OF LIGHT
19.1 Introduction
Can you list some of your everyday experiences that are based on the wave properties
of light? The colors observed in a soap bubble or an oil film on water are some
examples. What wave property of light is involved in these observations? Polaroid sun
glasses are designed to reduce glare from reflected light. What wave property of light is
used in these glasses? The laser is a new light source that has unique properties; can
you name some of these properties?
In this chapter we will explore the experimental basis of the wave phenomena of
light. We will discuss the questions mentioned here and point out some applications of
the wave nature of light.
19.2 Light Waves
Light is another example of an interaction-at-a-distance. Once again, as we discussed in
Chapter 2, we postulate that the source of light is the source of a field that fills the space
between the source and the receiver. We know of properties of light that can be
explained as interference or diffraction phenomena. Hence, we construct a model of
light as a wave. We shall begin by introducing you to the properties of our wave model
of light, and then we shall discuss how this model helps in understanding the various
properties of light. In our model light is a transverse electromagnetic wave. Light waves
consist of electric and magnetic fields perpendicular to each other oscillating with the
wave frequency at right angles to the direction of propagation. The speed of light in a
vacuum c is the maximum possible speed for the transmission of energy. (c ≅ 3.0 x 108
m/sec) The speed of light in matter is less than c. The index of refraction of a material is
c divided by the speed of light in the material. Since our model for light satisfies the
wave equation we have the following relationships:
c = fλ, n =c/v = fλo/ (fλ) = λo / λ
(19.1)
where λ is the wavelength in medium, n is the index of refraction of the medium, f is the
frequency of the light wave, and λ 0 is its wavelength in a vacuum. The traditional units
for light wavelengths are angstroms ( ), 1 = 10-10 m, and millimicrons (mµ), 1 mµ =
10-9 m. However in the approved SI, nanometers (1 nm = 10-9 m) are the proper unit to
use. The index of refraction for various types of glass is shown in Table 19.1 as a
function of the wavelength. The change in the refractive index of a glass with
wavelength is called dispersion and shows that light of different wavelengths travels
with different speeds in glass. Such glasses are called dispersive media.
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19.3 Interference
As we pointed out in our discussion of traveling waves, the superposition principle
applied to the wave model produces unique interference phenomena. For waves on a
string we found a complete range of possibilities ranging from total constructive
interference to total destructive interference. The determining factors were the
amplitudes and phase relationships among the waves present at a given place and time.
It was assumed that the waves were of the same frequency with constant phase
differences between any two waves. If the waves had been of different frequencies, the
phase difference between two waves would be continuously varying in time (except for
the special case where the frequencies were all multiples of a common frequency). What
are the necessary conditions for interference phenomena in light waves? What examples
of light interference have you observed in natural phenomena? What results when two
or more beams of light are present in the same region of space? The answer to this
question depends on the nature of the light beams. If the beams are from different
sources (or even different parts of a large single source), the resultant energy at any
point is the sum of the energies produced by the individual beams. Such independent
sources are called noncoherent sources. The law that governs noncoherent sources is
called photometric summation. If the beams are coherent (that is, the beams have a
constant phase relationship during observation), the results of the two or more beams
are the summation of the amplitudes of the individual beams. When the superposition
principle is applied to coherent light beams, the light intensity varies from place to
place, giving rise to intensity maxima and minima called interference fringes. The
maxima occur where the waves are in phase. The superposition principle leads to
constructive interference with the resultant amplitude equal to the sum of the in-phase
amplitudes. The minima (destructive interference) occur in places where the waves are
exactly out of phase (phase differences of one-half wavelength). A graphic example of
constructive and destructive interference due to coherent sinusoidal waves of unequal
amplitude is shown in Figure 19.1. What would you expect if the amplitudes were
equal in these two cases? How can you reconcile interference fringes with the principle
of conservation of energy?
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Figure 19.1 Summation of two waves. (a) In phase (constructive interference) and (b) out of phase (destructive interference). If you wanted to produce interference fringes, how would you do it? Thomas Young
was the first to record such an experiment, and it was accepted as the crucial
experiment in support of the wave model for light. The essentials of Young's
experimental set up are shown in Figure 19.2 .
Figure 19.2 (a) Diagram of experimental set up for double-­‐slit interference. (b) line drawing for double-­‐slit interference showing the path difference Δ as a function of the angle θ. The two slits have a width w and are a distance d apart. The viewing screen is a distance L from the slits. The distance along the screen is x. A narrow slit in front of the source provides a coherent plane wave that is incident
upon two very narrow slits. Each of these slits acts as a new source, but these two
sources are coherent, spatially separated sources. Every point on the screen represents a
particular optical path length (defined as the path from each slit to the screen along each
straight line). Since the slits are coherent sources with zero phase difference, we should
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expect constructive interference where the optical path difference between the two slits
is an integral number of waves.
Δ = mλ
(19.2)
where Δ is the optical path difference and m = 0, 1, ... From the geometry of the doubleslit apparatus (Figure 19.2b), we see that this relation can be expressed as:
path difference = Δ = d sin θ
where d = slit separation.
Thus, constructive interference occurs when
Δ = mλ = d sin θ
(19.3)
where m = 0, 1, 2, ... and is called the order number.
The zeroth order bright fringe is symmetrically centered with respect to the slits. There
is a first-order maximum on each side of the central bright fringe at an angle θ given by sinθ =λ/d. For small angles, sin θ ≅ tan θ, and sin θ ≅ x/L = mλ/d for bright fringes.
EXAMPLE
Given a pair of slits separated by 0.2 mm, illuminated by green light (500 nm) in a
coherent parallel beam, find the separation of the two first-order green fringes on a
screen 1 m from the slits.
The waves from the two slits must be in phase at positions of maximum intensity.
See Figure 19.3. For m = 1 (first order),
d = 2 x 10-4 m, λo = 5 x 10-7 m
sin θ = λ / d
sin θ = 5 x 10-7 / 2 x 10-4 = 2.5 x10-3
Figure 19.3.
For small angles,
x/L = tan θ ≅ sinθ ≅ θ
when θ is in radians.
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We want to find two times the distance between central bright and the first maxima
on either side of the central bright fringe,
2x = 2 x 102 cm x 2.5 x 10-3
x = 5 x 10-1 cm = 5 mm
19.4 Effective Optical Path Lengths
Young's experiments show that it is possible to set up interference fringes by
introducing optical path differences between coherent wave trains. In Young's
experiments the path differences were in air, and we did not correct for the difference of
the wavelength in a vacuum and the wavelength in air. (If n = 1.0003 for air, what is the
error in neglecting this wavelength change?)
Effective optical path lengths in materials involve the wavelength of white light in
the medium. For example, the path differences for constructive interference must be an
integral number of waves in the given material. Since the wavelength of light in a
material with an index of refraction n is given by its wavelength in a vacuum λ0 divided
by n, then a thickness t of this material is equivalent to nt/λ0 wavelengths. This means
that the effective optical path length for a sample of thickness t with an index of refraction n
is nt:
optical path length = nt
(19.4)
EXAMPLE
Compare the effective optical path of 10-6 m in a vacuum, air and water. 10-6 m (vacuum)
= 1.0003 x 10-6 m (air) = 1.3333 x 10-6 m (water)
19.5 Thin-Film Interference Patterns
Thin-film interference patterns are the result of the superposition of waves reflected
from the front surface and the back surface of the film. The phase difference between
these front and back surface reflected waves determines the nature of the interference
pattern for a given film system. An important factor that must be included in the
analysis is that light reflected at an interface where n2 > n1 (the second medium has a
greater index of refraction than the first medium) undergoes a one-half wavelength
phase shift (π radians) upon reflection. Light reflected from an interface wheren1 > n2
undergoes no phase shift.
Let us consider an oil film (n = 1.50) floating on water ( n = 1.33). When such a film is
viewed in white light, a series of colored fringes will be observed. We will calculate the
minimum thickness of the oil film for constructive interference of purple (400 nm) and
red (700 nm) light. There is a one-half wavelength shift at the air-oil interface and no
phase shift at the oil-water interface (Figure 19.4). Thus for constructive interference
between the first and second surface reflections, we must have a minimum optical path
of one-half wavelength in the oil. Note that any optical path length equal to an odd
multiple of one-half wavelength in oil will produce constructive interference for a given
wavelength, and an even multiple of one-half wavelength will produce destructive
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interference. The effective optical path difference between the front and back surface
reflections is 2tnoil, where t is the film thickness. Light reflected from the second surface
travels a distance of 2t in the oil. Thus the condition for constructive interference can be
expressed as
2tn = (m/2)λo
(19.5)
for m = 1, 3, 5, ... (constructive) and m = 2, 4, 6, ... (destructive), where t = film thickness
andn is the index of refraction of the film, or
t(purple) = λo/4n = 400 nm/6 = 66.7 nm
t(red) = 700 nm /6 = 116 nm
Consider the similar problem for an oil film suspended on a wire frame in air.
The colors of some insect wings are due to such interference phenomena. (What
information about insect wings could you obtain by studying these color fringes? What
experiments would you do?)
19.6 The Interference Microscope
The interference microscope is designed to convert the phase difference introduced by
the different optical path length through the specimen and through the surrounding
fluid into a difference in intensity that can be detected by the observer. This is done by
taking a parallel coherent beam that passes only through the air and making it π radians
(1/2 wavelength) out of phase with the beam passing through the fluid alone. These
two beams produce destructive interference on a view screen. Any light passing
through the specimen will introduce an additional phase shift. The phase shift Δθ is
given by
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421
(19.6)
where ns = specimen index and n f = fluid index. At places where Δθ = 2π there will be
constructive interference and as the specimen thickness varies, intensity variations
occur that make the specimen visible on the screen.
19.7 Diffraction
The bending of waves around objects is common to all wave motion. This wave
phenomenon is called diffraction. Diffraction patterns result from the interference of
waves that travel different distances around objects or through apertures. Consider the
diffraction due to a single slit as shown in Figure 19.5a .
To derive the conditions for destructive interference we divide the slit of width w into
two equal zones (see Figure 19.5b). For destructive interference the waves from zone A
cancel the waves from zone B in pairs. The path difference for these cancelling pairs
results in the following condition for first-order minima: (w/2) sinθ = λ/2 for the firstorder minima on either side of the central bright band.
Higher-order minima are given by path differences equal to odd multiples of onehalf wavelength. This general condition for single- slit diffraction minima can be
expressed as
w sin θ = mλ
where m = 1,3,5, ... is the order of the minimum and w is the width of the slit.
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EXAMPLE
Find the separation of the two second-order minima for red (600 nm) parallel light
incident on a slit of 0.100 mm width which is 1.00 m from the viewing screen.
w sin θ = 2λ
sin θ = 2 x 6.00 x 10-7 m/10-4 = 12.0 x 10-3
Let x2 be the distance along the screen from the central maximum to the second order
minimum, for small angles
sin θ ≅ x2 / L
so
x2 = 12.0 x 10-3 m
and the separation between the two second-order minima is 2x2 = 24 x 10-3 m = 0.024 m.
19.8 Diffraction Grating
The diffraction grating consists of a number of close, uniformly spaced, diffracting
elements (either transmitting slits or reflecting grooves). Such gratings are used in
spectrometers to measure the wavelengths of spectra. The diffraction pattern produced
by a granting is the result of the interference of the waves from the different diffracting
elements. A diagram of a diffracting grating is shown in Figure 19.6. Constructive
interference occurs when the waves from adjacent elements have path differences of an
integral number of wavelengths between the grating and screen. The equation
expressing this condition is:
d sin θ = mλ
(19.3)
where m = 1, 2, 3,... and is the order of the diffraction maxima, where d is the distance
between the slits and is considered to be much larger than the width of each slit.
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Figure 19.6 A diffraction grating, a screen with many fine slits equally spaced a distance d apart. EXAMPLE
Find the angular spread of the first-order visible spectrum from a grating with a
number of lines per unit length N of 10,000 lines/cm. The visible spectrum range is
from 400 nm (purple) to 700 nm (red).
d = 1/N = 10-4 cm
sin θ (purple) = 4 x 10-5 cm/10-4 cm = 0.4
θ (purple) = 23.6°
sin θ (red) = 7 x 10-5 cm/10-4 cm = 0.7
θ (red) = 44.6°
Δθ = θ (red) - θ (purple) = 21° angular spread of first-order visible spectrum
19.9 Resolution Factors
The ability of a grating to separate two wavelengths increases as the order increases.
However, there are problems concerning intensity and orders that overlap in high-order
spectra. The intensity of the diffraction decreases with increasing order. In addition, the
overlap for a given grating produces a higher-order, shorter- wavelength line between
lower-order, longer-wavelength lines in the spectra under observation. For example,
third-order 400-nm light will be at same angle as second-order 600-nm light. Thus, the
use of higher orders to improve resolution involves a compromise to solve the
resolution problem. We have ignored the single slit patterns that are superimposed for
the actual grating. Our treatment has assumed that the individual slits have widths
approximately equal to the wavelength of light.
The limiting factor in the resolving power of an optical instrument is the diffraction
pattern produced by the system. The apertures of optical instruments produce
diffraction patterns of point sources. The resolving power of an optical system is
measured by the minimum angle subtended by two just resolvable sources of light. The
"just resolvable" criterion we use is called the Rayleigh criterion. Lord Rayleigh
suggested that two images are just resolvable if the central maximum of one source is
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found at the first minimum of the other source. This condition leads to the following
equation for the resolving power of circular apertures:
sin θR = 1.22λ/D
(19.8)
where λ = wavelength, D = diameter of aperture, and θR is the angle between two just
resolvable sources. We see that the resolving power may be increased by using shorterwavelength light and/or larger apertures.
The resolving power of a telescope is illustrated in Figure 19.7. The images of just
resolvable stars are displayed on a screen in the focal plane of the telescope eyepiece.
The stars, separated by a distance s, are a distance L from the objective lens of the
telescope. The angle subtended by these stars is equal to the Rayleigh criterion angle for
light of wavelength λ and an objective of diameter D.
θR ≅ sin θR = 1.22λ/D = s/L
The distance x between the just resolvable images at the focal length of the eyepiece is
given by
x = fe θR = fe (1.22λ/D)
where fe is the focal length of the eyepiece of the telescope. The intensity pattern for
such resolvable images is shown in Figure 19.8 .
Figure 19.8 Intensity plot of Rayleigh’s criteria for resolution. Chapter 19- Wave Properties of Light
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19.10 Polarization
Wave amplitude oscillations in a plane perpendicular to the direction of propagation
are possible only for transverse waves. The orientations of the wave amplitude
vibrations in a plane perpendicular to the direction of propagation are called the
directions of polarization of the wave Figure 19.9. The transverse nature of light, with
its electromagnetic field oscillations perpendicular to the direction of propagation,
makes it possible to observe light polarization phenomena. In general, light sources
produce light waves that are randomly polarized. This means the electric field vector
(sometimes called the light vector) in the electromagnetic wave is vibrating with a
random orientation in a plane perpendicular to the direction of propagation (Figure
19.9) . Why do you think most natural sources produce unpolarized light?
Figure 19.9 Graphical representation of ordinary light showing the electric and magnetic wave amplitudes in planes perpendicular to the direction of the propagation of the light. There are three types of polarizations possible for the electric field vector in xy plane:
plane polarized, elliptically polarized, and circularly polarized. These three
polarizations are illustrated in Figure 19.10. The plane polarized wave can be resolved
into x- and y -components vibrating in phase with each other. The elliptical polarization
can be resolved into x- and y-components not in phase with each other. The circular
polarization can be resolved into x- and y-components 90ø out of phase with each other.
Circular polarization is said to be right handed if the electric field vector rotates in the
clockwise direction as you look at the wave as it travels toward you. Left-handed
circularly polarized light corresponds to the E vector rotation in the counter-clockwise
direction as the wave travels toward you. We can write the following rules for
combining two simple harmonic vibrations of same frequency and amplitude
perpendicular to each other:
1. When the phase difference is an even multiple of π, the result is plane polarization at
45° to both original vibrators.
2. When the phase difference is an odd multiple of π, plane polarization results at 90° to
those of case 1.
3. When phase differences are odd multiples of π/2, circular polarization results.
4. All other phase differences produce elliptical polarizations (see Figure 19.10d).
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Polarized light can be produced by double-refracting crystals such as tourmaline or
other polarizing materials which have an optical axis. Incident light will have all
vibrations except those along the optical axis of the polarizing material heavily damped.
That is, light other than that with vibrations parallel to the axis is absorbed. In this way
incident unpolarized light will be plane polarized when it leaves such a polarizer. The
amplitude of a plane polarized beam making an angleθ with respect to the optical axis
of the polarizer will be proportional to the component parallel to the other polarizing
optical axis, that is, amplitude ∝ cos θ. The intensity of the light wave is proportional to
its amplitude squared, thus
I = Io cos2 θ
(19.10)
where I0 is the intensity for θ = 0°. By using two polarizers you can set up a system in
which you can continuously change the output intensity by varying the angle between
the two polarizers. See Figure 19.11 .
Figure 19.11 Polarizer and analyzer for plain polarized light. Other ways to produce polarized light are by the reflection of light from a dielectric
surface or by scattering light from small particles. The boundary value problem of
electromagnetic waves incident upon a dielectric surface yields solutions showing that
the reflected light has various amounts of polarization. There is a critical angle of
incidence, called Brewster's angle, that results in a completely plane polarized reflected
wave. The condition for this critical angle is that the angle of incidence plus the angle of
refraction equals 90°, i + r = 90°. When this condition is imposed on Snell's law, we get
the equation
n1 sin i = n2 sin r
But
sin (90° - i ) = cos i
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Thus,
sin i/cos i = tan θB =n21
(19.11)
where n21 is the relative index of refraction of the two media and equals n2/n1, and θB is
the Brewster angle. This is known as Brewster's law, and at the Brewster angle of
incidence, the reflected polarized light will have its electric field plane polarized parallel
to the plane of the reflecting surface. Polarization by reflection at the Brewster angle is
illustrated in Figure 19.12.
One effect of polarization due to reflection is to produce enhanced polarization of
sunlight over a water surface. It is not surprising to find evidence that some water
insects have sensitivity to polarized light as this may be useful in survival in their
natural environment.
Light scattered from small particles is also selectively plane polarized. You can
detect this polarization in sky light. It is thought that bees use this light polarization in
their navigation from hive to food and back. Polarization by scattering occurs because
light is a transverse wave, and thus the particles will selectively scatter light polarized
perpendicularly to the direction of travel of the incident light. The particles of dust, ice,
and salt crystals in the atmosphere produce partial polarization of skylight due to
scattering. This same scattering, called Rayleigh scattering, also produces the blue color
of the sky. The intensity of the scattered light is proportional to 1/λ4, and thus the blue
region of the solar spectrum is scattered most, producing the blue sky. Can you use this
information to explain why a sunset is usually red? See Figure 19.13.
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There are many materials that have different indices of refraction for different planes
of polarization. This property of having different velocities for the propagation of light
of different polarizations is called double refraction, or birefringence. Calcite, for
example, is a birefringent crystal. For these uniaxial materials the two different
refractive indices are called ordinary and extraordinary, and each is slightly wavelength
dependent Table 19.2.
EXAMPLE
Given the index of refraction for an ordinary beam to be 1.320 and the index of
refraction of 1.330 for the extraordinary beam in a biological specimen (for a 400 nm
light), find the phase difference between these beams after passing through a specimen
1µ thick. t = 10-6 m, λ0 = 400 nm = 4 x 10-7 m
phase difference = 2π (optical path difference) /λ
phase difference = 2π (nc -no)t / λo
= 2π x 10-6 m x (1.330 - 1.320) / 4 x 10-7 m = π/20
Recall that optical path length = t n since λ =λo/ n.
19.11 Applications of Polarization
Since many biological specimens are birefringent, the phase difference introduced in the
specimen can be used in designing special microscopes. One such instrument is the
polarizing microscope which is used to increase the contrast between different parts of
the specimen under study. In some cases the actual thickness of the various parts can be
determined. The polarizing microscope makes use of polarized light transmitted
through the specimen. An analyzer is placed between the objective lens and the
eyepiece of the microscope. This analyzer is crossed (at 90°) with the polarizer. If the
object is homogenous throughout, the field of view will be dark. A different thickness,
or orientation, of the sample will produce a varying light intensity, and thus make
sharp contrasts visible within the sample. By using a compensating wedge of a known
birefringent material (usually quartz), this contrast can be enhanced for thin samples. A
calibrated wedge makes it possible to measure specimen thickness by cancelling out the
phase shift of the specimen with a known thickness of wedge.
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19.12 Optically Active Materials
Some substances (especially in living systems) have the property which causes the
plane of polarization of light to be rotated as it passes through them. Such materials are
called optically active materials. There are both right- and left-handed substances that
rotate the plane of polarization in opposite directions. Sugar is found in both forms,
right-handed dextrose and left-handed levulose. The optical activity is demonstrated by
solutions of these materials. The optical activity of a solution is proportional to the
concentration of optically active molecules in the solution and the length of the light
path through the solution. The angle of rotation of plane of polarization θ is given by
θ = KLc
(19.12)
where K = proportionality constant for a given substance, L = path length, and c =
concentration of optically active material.
This equation can be used with appropriate instruments such as a polarimeter or
saccharimeter to measure unknown concentrations of optically active substances.
Polarized light is passed through the sample and the angle through which the analyzer
must be rotated to achieve extinction is measured.
EXAMPLE
A concentration of dextrose of 1.00 g/cm3 produces a rotation of 5.3°/cm of path. If
light passing through 10.0 cm of an unknown dextrose solution is rotated through 8.3
degrees, find the concentration of the unknown. If 5.3° = K x 1 (cm) x 1.00 (g/cm3), and
8.3° = K x 10.0 (cm) x c (g/cm3) then
5.3 / 8.3 = 1.00 (gm/cm3) / 10.0 c or c = 8.30 / 53.0 = 0.156 g/cm3
19.13 Lasers and Laser Applications
In general, the light from a single light source consists of a series of random wave trains.
Each wave train lasts for about 10-9 sec, and thus a wave train is approximately 30 cm
long. Since these wave trains are uncorrelated with each other in phase, they are
noncoherent. Light from different sources is also noncoherent. The development of the
laser has provided a high intensity, monochromatic, highly directional, coherent source
of light.
The laser produces a continuous train of waves. The actual wave train from a laser is
approximately 30 m long. The laser beam is highly correlated in space and time. Laser is
an acronym for light amplified by the stimulated emission of radiation. There are many
different kinds of lasers available today, but the physical basis of each is essentially the
same. We will outline the basic physics involved in the operation of the laser in the
chapter on atomic physics (Chapter 28).
There are many applications of lasers because of their unique properties as light
sources. Their high intensity and extreme directionality has made them ideal for
alignment and communication applications. The laser beam can be focused by optical
lenses. It is possible to produce power concentrations greater than 109 watts/cm3 at the
focal point of a lens. Such power levels enable lasers to be used for cutting materials (for
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example, in laser band saws) and drilling small holes with great precision.This wellfocused high power beam has been used in medicine for such things as welding
detached retinas and bloodless surgery using the self-cauterizing property of the laser
beam. Recently the laser has been used to treat patients suffering from diabetic
retinopathy. This condition arises in diabetics when tiny blood vessels deteriorate and
new vessels grow on the surface of the retina. When these vessels hemorrhage into the
normally clear vitreous humor, vision is severely impaired. Scar tissue may also detach
the retina from the back of the eye. A fine laser beam focused on the weakened blood
vessels can produce coagulation and the proliferating new vessels can be destroyed.
The incidence of vision loss has been cut by 60 percent through use of the laser in this
treatment.
A similar application of the laser is the control of hemorrhaging in the
gastrointestinal tract. A fiber optic bundle can be inserted through the mouth into the
stomach. With the bleeding sites visible, the laser can be discharged through the bundle.
The intense laser beam can bring on coagulation and cease hemorrhaging within ten
minutes.
A new use of the laser is being investigated in dentistry. New tooth filling materials
that can be cured by laser radiation are being tested. It is hoped that a material that can
be welded to the tooth with a pulse of laser radiation will be found.
One of the most promising uses in medicine involves the laser in holography. The
hologram is a three-dimensional photograph of the exposed object. This photograph can
be used to reconstruct a three dimensional image of the original object. A diagram of a
set up used for producing holograms is illustrated in Figure 19.14.
The partially silvered mirror produces two coherent beams, a reference beam and a
sample beam. The sample beam is reflected from the object onto the film where it
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interferes with the reference beam producing an interference pattern on the film. This
interference pattern contains all of the three-dimensional information available from the
object. When a laser beam strikes this record at the angle of the reference beam, it
produces a three-dimensional image of the original object. Much research has been
devoted to holography, and it offers great potential in many applications. A threedimensional hologram television picture would be most useful for producing threedimensional images of organs inside of the body.
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Sandbox holography. (a) Set up for songle beam transmission hologram. The light source is a 3-­‐
milliwatt helium-­‐neon laser. The laser beam is incident upon a mirror which reflects part of the beam directly to the photographic film, and another part of the beam is reflected from the object (in this case a toy airplane) to the film. Thin film is placed between the glass plates held together by spring clamps. (b) A hologram taken with this set up. (Courtesy of Jeseph Ferry and Dr. Richar Anderson, University of Missouri-­‐Rolla.) 19.14 Light Interactions with Living Systems
The interactions of light and living matter can be classified as ionizing interactions and
nonionizing interactions. The ionizing radiation consists of short wavelength radiation
including ultraviolet, x-rays, and gamma rays. Of these we will consider only the
ultraviolet radiation in this chapter.
Ultraviolet radiation is important in photochemical reactions. The ultraviolet light
provides needed activation energy that makes certain reactions possible. Photosynthesis
in plants is a most important ultraviolet induced photochemical reaction.
Light can act as a catalyst in biochemical reactions. The synthesis of vitamin D in our
bodies is an example of a photocatalytic reaction. These reactions are the basis of much
ultraviolet therapy. Some people also experience reactions like allergies to ultraviolet
light. Some drugs (especially antibiotics) can induce an allergic sensitivity to ultraviolet
light. Such drugs combined with ultraviolet light treatments are being developed to
treat such chronic skin diseases as psoriasis. Prolonged exposure to ultraviolet light
(like sun bathing) can damage skin pigments. Proteins and pigments in the skin absorb
maximally at about 280 nm. Fortunately much of the sun's ultraviolet radiation is
absorbed by the ozone in the earth's atmosphere. There is currently concern that the
propellant gas in aerosol cans may be causing depletion of the earth's ozone layer. An
intensive research effort is underway to investigate this newly discovered
environmental threat.
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Ultraviolet light is used in treating jaundice in newborn babies. The condition
develops because of poor liver functioning in newly born babies. It has been found that
ultraviolet light can be used as a treatment until the baby matures and its liver functions
properly. Ultraviolet radiation is also used to kill bacteria. Special lamps with high
ultraviolet output are sold as bactericidal lamps.
Recent studies have shown a dramatic effect of ultraviolet light combined with a dye
(hematoporphyrin) that accumulates preferentially in malignant tissue. Studies with
rats have given kill rates of 75 to 90 percent for malignant brain tumors. This
interaction, like others between ultraviolet light and tissue, is not completely
understood. There remains much to learn about the interaction of ultraviolet radiation
and living matter.
Infrared light is a form of nonionizing radiation. Infrared radiation is associated
with the heating effects in molecular systems. The absorption of infrared light increases
the energy of molecules in the absorbing material. When infrared radiation penetrates
tissue, the associated warming effect is the basis of infrared therapy.
The visible spectrum (400-700 nm) is that part of the solar spectrum that is used in
human vision. This important electromagnetic interaction with our environment is the
subject of a separate chapter (Chapter 20). Another effect associated with visible light is
the entrainment of the individual's daily temperature cycle. Changing the light cycle for
an individual changes the phase of the body temperature cycle.
Photobiology offers an exciting area for research by biologists, chemists, and
physicists. There are many unanswered questions concerning the interaction of light
with living matter.
EXAMPLE
It is known that approximately 14 x 10-13 joule is required to kill a single bacterium. Find
the kill rate ΔN/Δt for a 20 watt uv bacterial light if 20 percent of its output is lethal for
bacteria.
effective killing power = 0.2 x 20 J/sec = ΔN/Δt x 14 x 10-13 J/bacterium
Thus
ΔN/Δt = 2.9 x 1012 bacteria/sec
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SUMMARY
Use these questions to evaluate how well you have achieved the goals of this chapter.
The answers to these questions are given at the end of the summary with the number of
the section where you can find related content material.
Definitions
1. Assign the correct term to each of the following physical phenomenon:
a. color of a soap bubble
b. solar spectrum produced with a prism
c. intensity pattern of light after passing through a pin hole
d. two beams of laser light used in holography
e. light characteristic when reflected from a dielectric surface
f. the rotation of the plane of polarization by sugar solutions
Applications of Wave Properties
2. Thin-film colors involve the superposition of reflected waves. Phase differences
involved arise from ______ and ______.
3. A diffraction grating spectrometer designed for high resolving power should have a
_______ number of lines/mm.
4. The resolving power of a telescope is improved by increasing the size of ______ used.
5. Holography results in a two dimensional ______ on photographic film that produces
a ______ when viewed with laser light under the proper conditions.
Problems Involving Wave Properties
6. If a soap film appears bright when viewed in 600 nm light, find its minimum
thickness if n = 3/2.
7. A diffraction grating with N lines per meter is L meters from a screen. For wavelength
λ find the separation between first-and second-order fringes on the screen. Assume
L>>d.
8. Light reflected from a surface at 53° is noted to be plane polarized.
a. If you look at this reflected light through Polaroid sun glasses find the fraction of
this light reaching your eyes if you tilt your head 45° from the vertical.
b. Find the index of refraction of the dielectric material.
Optical Activity
9. Sketch a system that could be used to measure optical activity.
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Lasers
10. List three important characteristics of laser light.
Answers
1. a. interference (Section 19.3)
d. coherent (Section 19.13)
b. dispersion (Section 19.2)
e. polarization (Section 19.10)
c. diffraction (Section 19.7)
f. optical activity (Section 19.12)
2. reflection, optical path differences (Section 19.5)
3. large (Section 19.8)
4. objective lens (Section 19.9)
5. interference pattern, three- dimensional image (Section 19.13)
6. 100 nm (Section 19.5)
7. LNλ (Section 19.8)
8. 50 percent, n = 1.33 (Section 19.10)
9. source polarizer, sample, analyzer, detector (Section 19.12)
10. coherence, monochromicity, directionality, intensity (Section 19.13)
ALGORITHMIC PROBLEMS
Listed below are the important equations from this chapter. The problems following the
equations will help you learn to translate words into equations and to solve single
concept problems.
Equations
n = λo/ λ, n =c/v
(19.1)
d sin θ = mλ m = 1,2,3,...
(19.3)
optical path length = nt
(19.4)
2nt = m/2λo m = 1, 3, 5, ..., constructive, m = 2, 4, 6, ..., destructive
(19.5)
Δθ = 2π (ns -n f) t / λo
(19.6)
sin θR = 1.22λ / D
(19.8)
2
I = Io cos θ
(19.10)
tan θB = n21
(19.11)
θ = KLc
(19.12)
Problems
1. The wavelength of one of the lines in the emission spectrum of sodium is 589 nm is a
vacuum. What is its wavelength in heavy flint glass?
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2. A coherent parallel beam of the green light (546 nm) of mercury is incident upon a
pair of slits. The separation of the first-order interference pattern is 2 mm from the
central image on a screen 1 meter from the plane of the slits. What is the distance
between the slits? (Hint: sin θ ≅ θ.)
3. What is minimum thickness of an oil film on water that will give destructive
interference for 546-nm light by reflection from the surfaces of the film? n = 1.50.
4. What is the sine of the angle of diffraction for the second-order maxima for the 546
nm light of the mercury spectrum incident upon a diffraction grating with 5000
lines/cm?
5. What is Brewster's angle for water? m = 1.33.
6. A polarizer and an analyzer are set for maximum intensity of transmission. If the
analyzer is turned through 37°, what is the new intensity of transmission?
7. An optically active material of a given concentration c 0 and path length 10 cm
produces a rotation of 10° of the plane of polarization. What would be the
concentration of the same material that would produce the same angle of rotation
for an optical path length of 15 cm?
Answers
1. 357 nm
5. 53°
2. 0.273 mm
6. 0.64 I 0
3. 182 nm
7. 2/3 c 0
4. 0.546
EXERCISES
These exercises are designed to help you apply the ideas from one section to physical
situations. When appropriate the numerical answer is given in brackets at the end of the
exercise.
Section 19.2
1. Plot the index of refraction as a function of wavelength for higher dispersion crown
glass and for heavy flint glass. What is the physical meaning of the slope of the
curve? Compare the slopes at 400 and 600 nm. What is the index of refraction of each
for the 546-nm green light from mercury? [~ 1.523 at 546 nm, ~ 1.656 at 546 nm]
2. If the wavelength of the green line of mercury is 546 nm in a vacuum, what is it in
water? In heavy flint glass? [410 nm, 331 nm]
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Section 19.3
3. Two narrow slits are spaced 0.25 mm apart and are 60 cm from a screen. What is the
distance between the second and third bright lines of the inference pattern if the
source is the 546 nm light from mercury? [0.13 cm]
4. Given a double slit with a separation of 0.2 mm, find the separation of consecutive
bright fringes on a screen 1 m from the slits for red (600 nm) parallel light incident
on the slits. [0.3 cm]
Section 19.5
5. What is the minimum thickness of the film of a soap bubble with a refractive index of
1.33 if the film shows constructive interference for the reflection of the yellow
sodium light (589 nm) at normal incidence in air? [110 nm]
6. What is the minimum thickness of a plastic film (index of refraction 1.4) on your eye
glasses which will give destructive inference for the reflection of light of wavelength
560 nm? [200 nm]
Section 19.8
7. What is the wavelength of a line which is diffracted 20° in the first order for normal
incidence upon a transmission grating? What is the second-order diffraction angle
for this wavelength? Assume the grating has a ruling of 6000 lines/cm. [570 nm,
43.2°]
8. For orders greater than one, there is an overlap of orders in the visible spectrum from
a diffraction grating. What is the basic relationship that shows this? What thirdorder line coincides with the second-order line of 589-nm light? [393 nm]
Section 19.9
9. Find the resolving power of the 508-cm Mount Palomar telescope. Use 550 nm for the
wavelength of light. Find the separation of just resolvable objects near Jupiter (6.5 x
106 km from earth). [1.32 x 10-7 rad, 0.86 km]
PROBLEMS
Each of the following problems may involve more than one physical concept. When
appropriate, the answer is given in brackets at the end of the problem.
10. For a double refracting crystal such as calcite, the
geometry may be such that only one beam of
light is transmitted while the other is internally
reflected at the surface of a 600 crystal. What are
the largest and smallest incident angles that can
be used to separate the two beams by this
method? Assume a sodium source ( λ = 589 nm).
See Figure 19.15 . [26.9°, 40.2°]
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11. A beam of light from a sodium arc is
incident upon a 60ø heavy flint
prism at the proper angle to give the
minimum angle of deviation. (This
condition calls for ray parallel to the
prism base inside the prism; see
Figure 19.16 .) What is the value of
this angle? [51.2°]
12. A piece of quartz crystal has two parallel sides 1 mm apart. A beam of yellow
sodium light (λ = 589) is at normal incidence on one of the parallel sides of the
crystal. What is the minimum change in phase difference between the ordinary and
extraordinary rays in going through the crystal? [96 rad]
13. How does the diffraction angle for a given wavelength line vary as a function of slit
separation for a given order? How does the diffraction between two wavelengths,
say 546 nm and 589 nm, compare in first and second and in second and third orders?
What does this suggest to you in trying to resolve two lines of wavelengths close
together? This solution may not be practical. Why? Remember in nature you do not
get something for nothing!
14. What is the maximum diffraction order of a red light of 656 nm that you can get
with a grating of 6000 lines/cm? 600 lines/cm? [Second order, twenty-fifth order]
15. A beam of light is incident upon a salt solution (1.36), and the reflected beam is
completely polarized. What is the angle of refraction of the beam? [36.3°]
16. How can you combine two beams of plane polarized light to produce a beam of
circularly polarized light?
17. A soap film (n = 1.33) is displayed in a vertical wire loop. It is noticed that the film
has a dark reflection band at the top just before it breaks. If green light (500 nm) is
used, find the thickness of the film at the position of the first bright band. [94 nm]
18. A thin film of water (n = 1.33) is floating on glycerine (n = 1.47). Find the minimum
thickness of water that will produce constructive interference for reflected red (600
nm) light. [226 nm]
19. Find the minimum thickness of an oil film (n = 1.50) on water that will give
constructive interference for reflected red (600 nm) light. [100 nm]
20. Compare the effect of temperature on thin-film interference due to thermal
expansion and variation of index of refraction n. For benzene Δn/ Δt = -6 x 10-4/°C
and Δ(vol)/Δt = αV0 = (.24 x 10-3V0/°C)
21. A reflected green light (500 nm) is completely plane polarized at an angle of
incidence of 53° on an insect wing. There is also a dark band in this reflected green
light at the thinnest part of the wing. Find this minimum thickness, assuming it is
due to destructive interference for the green light when viewed in air. [188 nm]
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22. For objects that are not self illuminating the criterion for resolution is given in terms
of the radius of the first dark fringe of the diffraction of a circular aperture. The
diffraction fringe radius r is given by r = 0.61 λ/n sin i, where i is the angle
subtended by the aperture at the object, λ is the wavelength and n is the index of
refraction of the object space. Two objects are said to be resolved when the
separation of images is equal to the diffraction fringe radius.
a. Find the percent improvement in resolving power of a microscope that is obtained by
using an oil immersion lens system with n = 1.5.
b. Find the separation of just resolvable objects in green light (500 nm) if n sin i (called
the numerical aperture) is one. [33 percent, 305 nm]
23. A standard sugar solution (1 g/cm3) is found to rotate the plane of polarization of
light by 5.4° per cm of path length. A sugar sample of unknown concentration
rotates the plane of polarization through 10.6° in a 5 cm long sample tube. Find the
concentration of the sample. [0.4 g/cm3]
24. The Brewster angle for a specimen is 55° for 500-nm light. If the specimen is 0.001
mm thick, find the phase shift introduced by this specimen as compared with an
equal thickness of water. [72°]
25. A thin wedge of air is formed between a sheet of glass 5 cm long and a horizontal
glass plate. One end of the sheet of glass is in contact with a glass plate. The other
end is supported by a thin metal film 0.05 mm thick. The horizontal plate is
illuminated from above with light 589 nm. How many dark interference fringes are
observed per cm in the reflected light? [17 fringes/cm]
26. Given a single slit of width D =kλ, where k = constant, find the angle of the firstorder minimum for each of the following values of k: a. 1; b. 10; c. 100; d. 1000.
[a. 90° b. 5.7° c. 0.57° d. 0.0057°]
27. Given a diffraction grating with 5000 lines per cm, find the diffraction angles for
bluish-purple (400 nm) light in the first and second order. At which order does the
400 nm light overlap the reddish-orange light (600 nm)? [11.5°, 23.6°, third]
28. A helium-neon laser source produces a second-order spectrum for light of
wavelength 632.8 nm at an angle of 30° using a certain diffraction grating.
a. Find the angle for the first-order sodium yellow (λ = 589 nm).
b. Assume the grating is a reflection grating of 1-m focal length. Find the secondorder separation in the focal plane for the sodium doublet lines of λ equal to 589.0
and 589.6 mm. [a. 13.5° b. 0.7 mm]
29. The axes of a polarizer and an analyzer are oriented at 60° to each other.
a. If polarized light of intensity I is incident on the analyzer system, find the
intensity of the transmitted light.
b. If the incident light (intensity I )is plane polarized at an angle of 30° with respect
to the polarizer axis, find the intensity of the transmitted light. [a. 0.25 I ; b. 0.188 I ]
30. What is the minimum thickness of a water film on glass that will give destructive
interference for 546-nm light by reflection from surfaces of the film? [102.4 nm]
Chapter 19- Wave Properties of Light