Download lenses and mirrors

Where can you find lenses and mirrors? It is easy to think
of devices that rely on lenses for their operation—certainly
eyeglasses and contact lenses, but also telescopes,
microscopes, cameras, binoculars and various types of
image projectors. Mirrors, too, are found on the walls of
every household, as well as perched high in the corners of
many convenience stores, where their convex shape allows
the entire store to be viewed from one vantage point. But
did you know that lenses are used to direct the light of
streetlamps, traffic lights and car headlights and taillights?
Miniature lenses focus the laser light in a CD or DVD
player. And even smaller lenses, created by varying the
index of refraction of a disk rather than by shaping a
homogeneous piece of glass, are used with optical fiber
devices. In this chapter, you will learn how reflection and
refraction make lenses and mirrors work.
Underwater Building (Albert Yee)
Chapter 5
LENSES AND MIRRORS
5.1 INTRODUCTION TO LENSES
Lenses are one of the most common optical elements—just consider how
many people you know who wear eyeglasses or contact lenses. Lenses are also
among the simpler elements to understand, at least on an elementary level, since
they work by refraction.
Thin Lens Approximation
The lenses you are most likely to be familiar with are pieces of plastic
or glass with smooth concave or convex surfaces, such as the lens whose cross
section is shown in Figure 5.1. This lens bends light at both surfaces by
refraction because each of the lens surfaces is in contact with air. As shown in
the diagram, light bends at the first surface, travels a distance through the
material of the lens, then bends again at the second surface. That is, light
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LIGHT: Introduction to Optics and Photonics
Optical
axis
Rays are
assumed to
bend along this
line
Optical
axis
Figure 5.1 - Thick (top) and
thin (bottom) lens. The thin
lens is an approximation of a
real lens- we model the lens
by a plane in space where
the light rays appear to bend.
enters and exits the lens at different distances from the optical axis, the
horizontal line through the center of the lens.
A thin lens represents an ideal situation in which light enters and exits
the lens at the same level (Figure 5.1). That is, light is assumed to bend at one
plane in space, and the fact that light travels through the lens is ignored. Thin
lenses are, then, approximations to real lenses. As you might imagine, the thin
lens approximation works better with lenses that have relatively flat curvature
and are thin at the center. Although the thin lens model is easy to understand,
it is usually not good enough for real-world applications.
Converging and Diverging Lenses
Consider a ray of light that is initially parallel to the optical axis in
Figure 5.1. As the ray passes through the lens it is refracted by the first surface,
travels in a straight-line path to the second surface, and is refracted again. We
will assume these are spherical surfaces, that is, each half of the lens may be
thought of as a slice cut from a sphere. If the radius of curvature is known for
each surface, Snell's law and geometry can predict the path of the ray. As you
can see in the top drawing of Figure 5.1, light is bent toward the normal as it
enters the lens at the left surface, and away from the normal as it leaves the lens
at the right surface.
If we apply Snell's law to several incident rays near the optical axis, the
rays will travel through the lens and then pass through a common point on the
right hand side of the lens. This point is called the focal point. (Later you will
learn why it is not really a "point.") The distance from the lens to the focal
point is the focal length. Remember that a thin lens is treated as a plane in
space, rather than a piece of real glass. When we speak of distances from the
lens we are referring to a distance measured from the plane in space where rays
bend, shown in Figure 5.1 as a dashed line through the center of the lens.
A lens that directs all of the incoming parallel rays to a focal point (as
in Figure 5.1) is called a converging, or positive lens. A lens that causes all the
incoming parallel rays to diverge, or spread out, after leaving the lens is called
a diverging or negative lens. The reason for the terms positive and negative
will be come clear when we calculate the focal lengths of these lenses.
Figure 5.2 shows the cross-sectional view of several types of
converging and diverging lens. Note that the converging lenses are thicker in
the center than at the edges and the diverging lenses are thinner at the center.
Each of these lenses has advantages and disadvantages in optical systems,
depending on the wavelength of light and how the lens is being used. For
example, plano-convex lenses can be used to minimize spherical aberration,
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Lenses and Mirrors
which we will discuss later in this chapter. Lens system design is a complex
topic, best reserved for more advanced treatments of optics. (See the references
at the end of this chapter.)
Biconvex
Biconcave
Plano Convex
Plano Concave
Positive Meniscus
Figure 5.2 – Cross sections of
converging, or positive, lenses
(top row) and diverging lenses,
or negative, lenses (bottom
row).
Negative Meniscus
\
Figure 5.3 shows the actual path of light rays from a laboratory light
source as they pass through plastic shapes that behave like two dimensional
converging and diverging lenses. In each case, parallel rays enter from the left
hand side of the photo and are refracted by the plastic shape. The top lens
converges the rays to a focal point, the point on the optical axis where the rays
cross. This is called a real focal point. The bottom lens causes the rays to
diverge as they leave the lens. In this case, you can follow the outgoing rays back
through the lens to find the point from which the rays seem to emerge. This is
called a virtual focal point.
You will meet the terms real and virtual again when we speak of the
images formed by lenses. In general, the description "real" indicates that rays
actually meet at a point in space. "Virtual" is used to describe the situation where
rays appear to diverge from a point.
5.2 CALCULATING FOCAL LENGTH: THE LENS MAKER'S
FORMULA
It would be very tedious to apply Snell’s law to each point on both
surfaces of a lens in order to find the focal length. Fortunately, a simple
equation (the Lens Maker's Formula, Equation 5.2) can be used to predict the
focal length of both converging and diverging lenses that have spherical
surfaces. The derivation of this equation makes a simplifying assumption,
called the paraxial approximation. It is important to understand the
implications of this approximation, because it means the use of the Lens
Maker's Formula is restricted to certain types of lenses and situations. Because
87
Figure 5.3 - Refraction of
light by converging (top)
and diverging (bottom) lens
shapes. The rays of light
enter from the left.
LIGHT: Introduction to Optics and Photonics
the paraxial approximation is based on the small angle approximation, which
we will frequently use in this text, we will take a closer look.
The Paraxial Approximation
c
r
s
b
θ
a
θ
r s
c
b
a
Figure 5.4 - The small
angle approximation. In
the triangle below, the
arc length s is
approximately the same
as the side b. Also, the
sides a and c are nearly
the same length as the
radius r.
As you know, Snell's law involves the sines of the angles of incidence
and refraction. It would greatly simplify the mathematics if we could replace
the sines of the angles with the angles themselves, so that Snell's law becomes
a linear equation
n1θ1 = n2θ2.
Replacing the sine by the angle (in radians) is called the small angle
approximation and it is used often in optics.
Consider the triangles of Figure 5.4 and recall the definition of the
angle in radians: θ = s/r. If θ is very small, as in the triangle at the bottom of
the figure, the hypotenuse c and the radius r are nearly the same size. Also, the
arc length s is approximately the same length as side b. Therefore,
sin ! =
b
c
"
s
r
=!
That is, if the angle is very small, the sine of the angle can be replaced by the
angle itself, expressed in radians.
A useful addition to the small angle approximation involves the tangent.
Because side a and the hypotenuse c are nearly equal if θ is very small,
tan ! =
b b
" = sin !
a c
To summarize, the small angle approximation may be written
Small Angle
Approximation
(5.1)
sin ! " tan ! " ! (in radians)
To see how the small angle approximation applies to a lens, consider
the two parallel rays incident on the curved surface shown in Figure 5.5. The
θ'
ray that strikes closer to the optical axis has a much smaller angle of incidence,
θ
so the small angle approximation is more valid near the axis. In fact, paraxial
means "near the axis."
When is the paraxial approximation valid? It depends, of course, on the
circumstances and the precision required. It also depends on the curvature of the
Figure 5.5 - The paraxial
lens surface and how large an angle the incident rays make with the optical axis.
approximation applies to rays
For educational quality lenses in a school lab, the approximation works quite
striking the lens near the
optical axis. In this diagram
well, especially for relatively long focal length lenses. It is not appropriate,
the angle θ is much smaller
however, in the design and construction of precision optical systems. Fortunately,
than the angle θ'.
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Lenses and Mirrors
computer software is available to assist in the solution of complex lens design
problems.
The Lens Maker's Formula
The Lens Maker's Formula is derived in the Appendix by applying
Snell's law to a spherical surface. We will simply state the result here, and show
how it can be used to calculate the focal length of a lens with spherical surfaces.
We assume that light enters the lens from the left, as in the illustrations of Figure
5.1. Let n be the index of refraction of the lens material and R1 and R2 be the radii
of curvature of the left hand side and right hand side of the lens, respectively. If
the lens is surrounded by air (nsurr=1), its focal length, f, can be calculated by
1
f
(
"
) $# R1
= n!1
1
!
%
'
R2 &
1
Lens Maker's Formula
(5.2)
When you work with lenses, it is very important to be consistent with the
signs of various quantities (positive or negative.) You have probably already
encountered this in math and perhaps physics courses. To use the Lens Maker's
Formula, we need a sign convention to distinguish between curvature opening
toward the left and toward the right. We will say a curvature is positive if the
open "c" curve points in the positive direction (to the right); otherwise it is
negative (see Figure 5.6). This sign convention is based on the familiar Cartesian
coordinate system. In general, negative quantities mean "to the left" of the origin
and positive quantities are "to the right" of the origin.
Positive radius of curvature: R>0
Negative radius of curvature: R<0
Figure 5.6 - Sign convention
for spherical surfaces.
EXAMPLE 5.1:
The lens shown below (shaded blue) has a radius of curvature on the left
side of +10 cm and on the right side of -40 cm. It is made of glass with an
index of refraction of 1.5 What is its focal length? (Hint: Make a drawing
of the lens to guide your solution.)
+10 cm radius
Solution
Using Equation 5.2:
1
f
Lens
1 %
" 1
-1
!
= 0.0625 cm
# 10cm !40 cm '&
= (1.5 ! 1) $
-40 cm radius
-1
f = 1/ 0.0625 cm = 16.0 cm
Note that in Example 5.1, the left side of the lens is curved so that it
opens toward the right; this is a positive radius of curvature according to our
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LIGHT: Introduction to Optics and Photonics
sign convention. Using the same sign convention, the right side of the lens has
a negative radius of curvature. Incoming parallel rays on the left side of the
lens will meet at a focal point 16 cm to the right side of the lens. The focal
length is positive; this is a converging lens and the focal point is described as
"real."
EXAMPLE 5.2:
The lens shown at left has a radius of curvature on the left side of +40 cm
+40 cm radius
and on the right side of +10 cm. It is made of glass with an index of
lens
refraction of 1.5. What is its focal length if it is used in air?
Solution
Using Equation 5.2,
1
+10 cm radius
f
1 %
" 1
-1
!
= !0.0375 cm
# 40 cm 10 cm '&
= (1.5 ! 1) $
f = 1/-0.0375 = -26.7 cm
The focal length is -26.7 cm; rays appear to diverge from point this far to the
left of the lens.
The diverging lens of Example 5.2 has a negative focal length.
Remember that negative quantities mean "to the left" of the origin, in this case,
the lens. The negative focal point is the point to the left of the lens from which
rays appear to diverge (see Figure 5.2). This is a virtual focal point.
Optical Power
Often, lenses are specified by their focal lengths expressed in millimeters
or centimeters. But it is also common to refer to the power of a lens, especially
in ophthalmic applications. The optical power of a lens in air is given by
Lens Power (in air)
(5.3)
P=
1
f
When the focal length is in meters, the unit of optical power is diopters
(D), where 1 diopter = 1 m-1. An eyeglass prescription that specifies spherical
power of -2.25 D is describing a lens that has a focal length of f = 1/(-2.25 D)
= -0.44 m = -44 cm. Like focal lengths, powers are positive for lenses that bring
light to a focus and negative for lenses that cause light to diverge.
Finally, in some applications we may want to focus light in only one
direction, creating a "line" focus rather than a point. In this case the axis, or
direction of the lens, is an important specification. In Figure 5.7, the axis is
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Lenses and Mirrors
vertical, and incoming parallel rays of light are brought to a focus along a vertical
line. Cylindrical lenses are used in eyeglasses to correct astigmatism,
irregularities of the cornea that cause vision to be blurred in one direction only.
In this case, the cornea has an elongated, rather than spherical, shape. Eyeglasses
are made with cylindrical lenses whose axis is 90o to the cornea's axis.
Cylindrical lenses are also used to reshape laser beams that have an elliptical
cross section, creating a more circular beam profile, and to create "line" sources
of light for specialized applications. Cylindrical power is also measured in
diopters.
5.3 IMAGE FORMATION WITH THIN LENSES
Ray Tracing
The focusing properties of a lens lead directly to the formation of
images. The location and size of an image can be determined by considering the
rays of light that leave an object located on one side of a lens, and following their
progress through the lens. Determining where an image will form by following
rays through an optical system is called ray tracing. Although we can find the
location of an image using algebra, ray tracing has the advantage of providing a
visual explanation of why an image forms where it does.
Consider an object on the optical axis in front of a converging lens and
well outside the focal point. (Note that in our drawings light will nearly always
go from left to right, so "in front of" means to the left.) The object is an arrow;
that is, it has an orientation in space so that we can determine if the image is in
the same orientation (upright) or upside down (inverted). Figure 5.8 illustrates
such an object and uses a double-headed arrow to represent the (thin) converging
lens. Two focal points are shown, one on each side of the lens, because incoming
parallel light from either direction will be focused the same distance from a thin
lens. In other words, if the thin lenses in Figure 5.2 were turned around they
would focus light at the same distance from the lens.
The object may be illuminated, or it may be diffusely reflecting light
from the sun or artificial source. In either case, rays of light are heading out in all
directions from every point on the arrow. It would be an overwhelming task to
follow each of these rays, instead three principle rays are drawn to determine the
image location. These rays are chosen because we know in advance how they
will behave when they pass through the lens. Of course, they aren't the only rays
that leave the object and end up forming the image; they are just the simplest rays
to draw. In Figure 5.8, the heavy (blue) arrows represent the principle rays and
the remaining arrows indicate a few of the other rays that leave the tip of the
object arrow and meet at the tip of the image arrow.
91
Figure 5.7 - A cylindrical
lens focuses in only one
direction.
LIGHT: Introduction to Optics and Photonics
The principle rays are:
•
The parallel ray enters the lens traveling parallel to the optical axis.
It leaves the lens and passes through the focal point on the opposite
side. This is a direct result of the definition of the focal point.
•
The focal ray enters the lens after passing through the focal point on
one side of the lens. (Depending on object location, it may leave the
object heading in a direction as if it came from the focal point.) The
focal ray exits the lens traveling parallel to the optical axis. A
moment's thought reveals that this ray is the parallel ray traveling in
the reverse direction (right to left in our drawing).
•
The central ray heads from the tip of the object (arrow) toward the
center of the lens. It can be shown to pass through the center of the
lens undeviated.
Thin converging lens
Figure 5.8 - Ray tracing to
locate an image. The lens is
represented by the double
headed arrow. The parallel,
focal and central rays are
traced from the object to a
point on the right side of the
lens, where they form a real
image. Note that the rays
continue beyond the image.
At other distances from the
lens, the image is not in
focus.
Object
Image
Parallel ray
Focal ray
Central ray
Focal points
Notice that the three principle rays all meet at a point beyond (to the right
of) the lens before continuing onward. In fact, for every point on the object,
there is a point beyond the lens where rays originating at that object point meet.
The image points form a real image. The light can be projected onto a screen
held at the image plane in space, and a real image will be seen. Note that the
image is inverted (upside down) with respect to the object. The ray diagram also
indicates that, for this case, the image is smaller than the object. Carefully drawn
ray diagrams can be used to find at least an approximate solution to many lens
problems.
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Lenses and Mirrors
EXAMPLE 5.3
An object is placed 10 cm in front of a diverging lens with a focal length of
-5 cm. Where does the image form?
Solution
The object and lens with its focal points are drawn to scale along the optical
axis. Notice that a diverging thin lens is represented by a line with inverted
arrowheads at each end.
The three principle rays are followed from the tip of the object through the
lens as shown in the diagram below.
•
The parallel ray exits the lens as if it came from the virtual focal
point on the left hand side (see Figure 5.2).
•
The focal ray heads toward the focal point on the opposite (right)
side of the lens and exits the lens parallel to the optical axis.
•
The central ray passes through the center of the lens undeviated.
These three rays do not meet at a common image point on the right hand
side of the lens. Instead, if the three rays are followed backward (to the left)
it appears as if they originated at a point on the left side of the lens. That is,
if you look into the lens from the right side you will see a small virtual
image to the left of the lens.
object
image
Parallel ray
Focal ray
Central ray
Focal points
Thin Lens Equation
Although ray tracing can be used to locate the image formed by a lens, it
can be a fairly time-consuming process. It would be handy to have a formula that
would allow us to solve lens problems algebraically rather than with ray tracing.
(Lens designers actually use sophisticated computer programs that perform the
calculations and provide graphical and numeric information to assist in
optimizing optical systems.)
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LIGHT: Introduction to Optics and Photonics
Before we can numerically analyze problems with lenses, we need to
introduce some terminology and agree on a sign convention for distances along
and at right angles to the optical axis. General physics textbooks often use a sign
convention referred to as the "empirical convention.” Unfortunately, its rules
are only easy to follow for the simplest of lens problems. In this book we will use
the Cartesian sign convention, which is the one preferred by optical engineers. It
has the advantage of following mathematical rules you already know from
graphing in a Cartesian coordinate system and it is easily extended to
complicated multiple lens systems.
Figure 5.9 shows the Cartesian convention we will use throughout this
text. We assume that light is traveling from the left toward the right. The arrow
indicating the direction of light’s travel also indicates the positive direction. That
means up and to the right are both positive directions, down and toward the left
are both negative directions. Note that the optical axis marks the zero in the
vertical direction. In the horizontal direction, the lens itself divides the axis
between positive and negative. This is the convention we have already used to
locate positive and negative focal points when using the Lens Maker's Formula.
+
Light from left to right
Figure 5.9 - Cartesian sign
convention for solving
problems with lenses.
Optical axis
+
EXAMPLE 5.5
What are the signs (+ or -) of the distances indicated on the drawing below?
Solution
The solid arrow on the left has a height of +5 cm and it is located -10 cm
from the lens. The striped arrow on the right has a height of -5 cm and it is
located +10 cm from the lens.
10 cm
5 cm
10 cm
94
5 cm
Lenses and Mirrors
Now that we have a consistent sign convention we can use the thin lens
equation to locate the image for a given lens and object. The derivation of this
equation can be found in the Appendix of this text. The thin lens equation
follows from the application of Snell's Law to a spherical surface.
Let us call the distance from the lens to the object do (the object distance)
and the distance from the lens to the image di (the image distance), as shown in
Figure 5.10. The thin lens equation relates the object and image distance to the
focal length of the lens.
1 1 1
+ =
d o f di
(5.4)
Thin Lens Equation
The thin lens equation is easy to remember by noting the order of the
denominators and comparing to the geometry of a typical imaging problem;
starting on the left, light goes from the object (do), through the lens (f), and forms
(=) an image (di ). The equation can be used for both converging (positive focal
length) and diverging (negative focal length) lenses. This form of the equation is
correct for the Cartesian sign convention; a different sign convention may require
the order of variables be changed.
Light from left to right
Lens
do
Figure 5.10 - Geometry of the
thin lens equation. In this (and
many common situations) the
object is drawn to the left of
the lens and the object
distance is negative.
di
Object
Image
Equation 5.4 also suggests a practical way to find the focal length of a
lens. Subtracting the quantity 1/do from both sides gives
1 1 1
= !
f di d o
(5.5)
For a converging lens, which forms a real image whose distance from the lens is
easily measured, the focal length can be determined by measuring the object and
image distances and solving Equation 5.5.
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LIGHT: Introduction to Optics and Photonics
EXAMPLE 5.5
An object is located 40 cm from a lens with a focal length of 20 cm. Find
the location of the image.
Solution
Assume the object is to the left of the lens, then the object distance is
-40 cm. Using Equation 5.4
1
1
1
+
=
-40 cm 20 cm di
1
= 0.025 cm !1
di
di = 40 cm
The image forms 40 cm to the right of the lens.
In Example 5.5, the object was well beyond the focal point of the lens,
on the left hand side. What happens if we bring the object in closer to the lens?
Table 5.1 shows the calculated image distances for the same 20 cm focal length
lens for object distances that vary from -500 cm to 10 cm. You can use the thin
lens equation to verify the values in the table.
Object distance
Image distance
-500 cm
21 cm
-200 cm
22 cm
-50 cm
33 cm
-40 cm
40 cm
-25 cm
100 cm
-22 cm
220 cm
-20 cm
no image
-10 cm
-20 cm
Table 5.1 Object and image distances for a positive lens with f = 20 cm.
Notice that when the object is very far from the lens, the image is near
the focal point. This is to be expected. Wavefronts from very distant objects are
nearly flat when they enter the lens, or, said another way, the rays entering the
lens are nearly parallel. Thus, very distant objects are imaged near the focal point
as shown in Figure 5.2. This gives another way to quickly determine the
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Lenses and Mirrors
approximate focal length of a converging lens. Find a very distant object, such as
a tree or utility pole, and use the lens to form an image of the object on a piece of
paper. The distance from the lens to the paper is the approximate focal length of
the lens.
As the object moves closer to the lens, Table 5.1 shows that the image
moves away from the focal point. If the object is exactly at the focal point, there
is no image formed at all. (The image distance is sometimes said to be "infinite.")
In fact, if a point of light is placed at the lens focus, the rays are made parallel by
the lens.
To interpret the final result in the table, a negative image distance, recall
that negative distances are on the left side of the lens—the side that light is
coming from. This is a virtual image, like the one formed by the diverging lens in
Example 5.3. Recall that a virtual image is formed on the right hand side of the
lens and the eye follows the rays back to see the image. To see this image you
need to look through the lens back toward the direction of the object. This is how
a positive lens is used as a magnifying glass.
Magnification
The thin lens equation provides information on the location of an image
produced by a lens. But what is the size of the image? The definition of
transverse magnification (that is, at right angles to the optical axis) is the ratio of
the image height to the object height
M=
hi
ho
(5.6)
Using the two similar triangles formed by the object, image and optical
axis in Figure 5.11, the magnification can also be shown to be related to di and do
M=
ho
hi di
=
ho do
(5.7)
di
do
hi
Object
Le
ns
Transverse Magnification
Figure 5.11 - Transverse
magnification is the image height
divided by object height.
Image
Note that magnification can be greater than one, meaning the image is
larger than the object, or less than one, indicating that the image is smaller than
the object. Also, magnification is a positive quantity if the image is upright, or in
the same orientation as the object, and a negative quantity if the image is inverted
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LIGHT: Introduction to Optics and Photonics
or "upside down" from the object. For Example 5.5, the object distance was
40 cm and the image formed at +40 cm. Magnification is therefore -1; the image
is the same size as the object, but inverted.
It should be clear that a diverging lens cannot produce a real image, since
it cannot cause rays to come to a focus. In problems with a single lens, the thin
lens equation used with a negative focal length lens will always produce a
negative image distance, that is, a negative lens produces a virtual image.
Examples 5.6 and 5.7 illustrate the method for finding image location
and magnification for a converging lens. Note that a diagram showing the
relative positions of the object and lens help you to predict the type of image that
will be formed. The procedure for a diverging lens is the same; the problems at
the end of this chapter provide an opportunity to work with both types of lenses.
EXAMPLE 5.6
An object is located 25 cm to the left of a lens with a focal length of 10 cm.
3 cm
f
The object is 3 cm high. Find the image distance, magnification, height,
f
orientation and type (real or virtual).
25 cm
Solution
Since the lens is converging and the object is beyond the focal point, we
expect the image will be real and inverted.
a. Image distance: Use Equation 5.5, the thin lens equation
1 1 1
+ =
do f di
1
1
1
+
=
!25cm 10cm di
di = 16.7 cm
(Note that the object distance is negative.)
The image forms at 16.7 cm to the right of the lens. (The image distance is a
positive quantity.) This is a real image, as expected.
b. To find magnification use Equation 5.7
M=
di
do
M=
16.7cm
= !0.67
!25cm
The image is 0.67 (2/3) the size of the object, so it is 2 cm high. The
negative sign means that the image is inverted (upside down) compared to
the object, as expected. In summary, the image is real, inverted, 2 cm high
and 16.7 cm to the right of the lens.
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Lenses and Mirrors
EXAMPLE 5.7
An object is located 5 cm to the left of a lens with a focal length of 10 cm.
The object is 3 cm high. Find the image distance, magnification, height,
orientation and type (real or virtual).
f
Solution
Since the object is between the focal point and the lens,
f
3 cm
5 cm
we expect a virtual, upright image.
a. Image distance: Substituting these values into the thin lens equation
1 1 1
+ =
d o f di
1
1
1
+
=
!5cm 10cm di
(Note that the object distance is negative.)
di = -10 cm
The image forms at 10 cm to the left of the lens. (The image distance is a
negative quantity.) This is a virtual image as expected, you need to look
through the lens to see the image.
b. Magnification
M=
di
do
M=
!10cm
=2
!5cm
The image is twice the size of the object, so it is 6 cm high. The
magnification is positive, which means that the image is upright (same
orientation) compared to the object, as expected. In summary, the image is
virtual, upright, 6 cm high and appears to be 10 cm to the left of the lens
when viewed looking back through the lens.
Finally, the thin lens equation also provides a means of calculating image
location and size for multiple lens systems. The image formed by the first lens
becomes the object for the next lens, and so on. In these cases, it is absolutely
necessary to begin with a careful drawing of the lens system!
5.4 REAL LENSES: ABERRATIONS
The thin lens equation works reasonably well for many applications not
requiring precision. But for exacting applications, the thickness of the lens
becomes an issue. Although the discussion of "thick" lenses is beyond the scope
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LIGHT: Introduction to Optics and Photonics
Figure 5.12 - Spherical
aberration. Rays that pass
through the outside
portions of the lens focus
closer to the lens than rays
that pass through the
center of the lens.
White light
Blue focus
Red focus
Figure 5.13 Chromatic
aberration causes
different wavelengths to
focus at different
distances from the lens.
of this text, we will discuss two of the many aberrations that degrade the images
formed by a lens.
Spherical aberration is a result of the spherical geometry of the lens
surfaces. Rays passing through the outer portions of a lens with spherical
surfaces do not focus at the same point as rays that pass through the center
(Figure 5.12). Instead of a sharp focus, the focal region is actually a small spot.
This means that any images made with the lens will be blurred to some degree. A
lens designed to correct for spherical aberration may include stops to restrict the
usable part of the lens to the center. Alternatively, the radius of each surface can
be chosen to minimize the problem; for example, the side that faces the object
may be considerably more curved than the other side. Another solution is to use
"aspherical" surfaces. These are more expensive to produce than spherical lenses.
Chromatic aberration is caused by dispersion, the variation of index of
refraction with wavelength. Chromatic aberration results in a slightly different
focal point for each wavelength, which causes colored fringes to appear around
images of black and white objects (Figure 5.13). Lenses that compensate for
chromatic aberration are called achromats. Achromatic doublets are compound
lenses consisting of two lenses, one converging and one diverging, made from
different glasses and cemented together.
5.5 MIRRORS
Mirrors work by specular reflection, and as you know, mirrors can form
images. The formation of an image in a plane mirror is shown in Figure 5.14,
using the law of reflection. Two of the rays from the top of the cylindrical object
are traced, showing the reflection from the mirror. The rays are redirected toward
the eye and the eye and brain interpret the diverging rays as originating from
behind the mirror, that is, the image is virtual. Simple geometry can be used to
show that the image distance is equal to the object distance in a plane mirror.
reflective
surface
Figure 5.14 Image formation in a
plane mirror.
do
di
Common household mirrors are constructed of glass with a metallic
coating on the back. The back coating makes it easy to clean the mirror and
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Lenses and Mirrors
protects the coating from dirt and fingerprints. Precision technology applications
demand front surface mirrors, however, since back surface mirrors produce
double reflections—one from the glass surface and one from the mirror surface
on the back. These mirrors may be thin metal films on a glass substrate, or they
may be constructed of very thin layers of dielectric (non-conducting) materials.
Mirrors may be broadband, reflecting a wide band of wavelengths equally, or
they may be "tuned" to reflect a narrow range of wavelengths while transmitting
all others. Thin film mirrors and filters will be discussed more fully in Chapter 6.
Spherical mirrors
As with a lens, image problems with mirrors can be solved by ray tracing
or algebraically. Again, before introducing the equation for solving spherical
mirror problems, we need to have a sign convention. Although the law of
reflection has a simpler form than the law of refraction, the sign convention for
image formation by a mirror is a bit less intuitive than that for lenses. Recall that
for lenses the direction of light propagation defines the positive direction. In
Figure 5.9, light is pictured going from the left to the right, so that all
measurements to the right of the lens are positive and those to the left of the lens
are negative. Light travels through the transparent lens, so positive and negative
directions remain fixed for the duration of the problem, as light goes from object
to lens to image.
The situation with a mirror is complicated because after striking the
mirror, light reverses direction and returns to the direction from which it came.
This means that in order to be consistent, the sign convention must change when
the direction of light changes. Figure 5.15 illustrates the two sign conventions
for before and after reflection from a mirror. The figure on the left shows light
traveling from left to right toward the mirror. This situation is identical to the
sign convention used in Figure 5.9 with a lens. In the figure on the left, light has
been reflected from the mirror and is traveling from right to left. Now the signs
along the optical axis are reversed and distances measured from the mirror to a
point to the left of the mirror are positive, and from the mirror to points to the
right of the mirror are negative.
Light from left to right
+
Light from right to left
+
+
-
mirror surface
+
-
mirror surface
-
101
Figure 5.15 Cartesian sign
convention for mirrors
leads to results
consistent with those
for lenses: real image
distance is positive
and virtual image
distance is negative
(behind the mirror).
LIGHT: Introduction to Optics and Photonics
Although this sign convention may seem confusing at first, it leads to
results consistent with those for lenses. Real images will have positive image
distances and virtual images will have negative image distances.
Consider a practical example. When you look at an object in a mirror you
see a virtual image "behind" the mirror (Figure 5.16). The object distance is
negative; the object is situated to the left of the mirror and light travels from left
to right. However, the image distance is also negative because, after reflection,
light is traveling away from the mirror from right to left. Image distances
measured from the mirror toward the right (where the virtual image is located)
are negative.
mirror
Figure 5.16 - Sign convention for a mirror
problem. The object distance is negative
because it is to the left of the mirror when
light is going from left to right. The image
distance is also negative because the light
that forms the image goes from right to left.
do
di
Light
Light
Before reflection:
Object distance < 0
Center of curvature
Focal point
Figure 5. - 17 A concave
spherical mirror has a
real (positive) focus
Mirror focal length
After reflection:
Image distance < 0
The Spherical Mirror Equation
Ray tracing with spherical mirrors is much like the process with lenses.
Figure 5.17 shows the cross section of a concave spherical mirror. The mirror
itself is a portion of the shell of a sphere; think of a "slice" from a hollow rubber
ball, silvered on either the inside (concave) or outside (convex) surface. The
radius of curvature of the mirror in Figure 5.17 is negative because the curve of
the mirror opens toward the left (following the sign convention of Figure 5.6).
Using geometry, it can be shown that parallel rays of light that strike this
mirror are brought to a focus at a point halfway to the center of curvature. That
is, the focal length is equal to one half the radius of curvature. In order to
preserve our Cartersian sign convention and to have a positive focal length for
this converging mirror, we write
(5.8)
f =!
R
2
It should be noted that, as with a lens with a spherical surface, the
incoming rays shown in Figure 5.17 will not all be focused at precisely the same
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Lenses and Mirrors
point. There will be spherical aberration with rays striking the outer portions of
the mirror focused closer to the mirror than rays striking the center. If the mirror
is relatively flat and only rays close to the axis are considered (the paraxial
approximation), spherical aberration may be ignored in some applications. Using
a parabolic surface eliminates spherical aberration .
The concave mirror shown in Figure 5.17 has a positive focal length and
is capable of bringing rays to a focal point. A convex mirror (with a positive
radius of curvature) will have a negative focal length and a virtual focal point
from which incoming parallel rays will appear to diverge. A diverging mirror,
like a diverging lens, cannot produce a real image.
As with lenses, it is possible to use three principle rays to determine
where images are formed by a spherical mirror:
• The parallel ray strikes the mirror after traveling parallel to the optical
axis. This ray reflects so that it goes through the focal point (or so that it
appears to come from the focal point for a convex mirror).
• The focal ray strikes the mirror after passing through the focal point. (For
a convex mirror, the ray cannot pass through the virtual focal point, but
is heading in that direction before striking the mirror.) This ray reflects
parallel to the optical axis.
• The central ray goes through the center of curvature of the mirror (or is
headed in that direction for a convex mirror). Because a ray that passes
through the center of curvature strikes a spherical surface normally, it
reflects back along the line of incidence.
Objec t
Focal point
Object
Center of
curvature
Real
image
Virtual
image
There are other similarities between mirrors and lenses. Figure 5.19
shows ray diagrams for a concave mirror with the object outside (to the left of)
the focal point and a concave mirror with the object inside (to the right of) the
focal point. When the object is to the left of the focal point, the three principle
rays meet at a point after reflecting from the mirror. That is, a real image is
formed which can be projected onto a screen. (Of course, the screen must be
103
focal point
center of curvature
Figure 5.18 – A convex
mirror has a virtual focus
(behind the mirror).
Figure 5.19 - Ray
diagrams for a concave
mirror. On the left, the
object located beyond
the focal point produces
a real image. On the
right, the object is
between the focal point
and the mirror, creating
a virtual image. Note the
similarities and
differences in ray
diagrams for lenses!
LIGHT: Introduction to Optics and Photonics
placed below the object so that it does not prevent the light from the object from
reaching the mirror.) Note that the real image is inverted; as with a lens, the
magnification depends on the exact object placement.
When the object is between the focal point and the mirror, the reflected
rays diverge as they leave the mirror so no real image is formed. However, if you
place your eye to intercept the diverging rays, a virtual image is formed at the
point from which the diverging rays appear to emanate. This image is upright
and enlarged. This is a how a shaving or make up mirror works to create an
enlarged view of your face.
A ray diagram for a convex mirror is shown in Figure 5.20. In the figure,
the ray that leaves the tip of the object traveling parallel to the optical axis (R1) is
reflected so that it appears to come from the virtual focal point (behind the
mirror). The ray that leaves the object heading toward the virtual focal point (R2)
reflects so that it is parallel to the optical axis. The third principle ray (R3) heads
toward the center of curvature (behind the mirror) and reflects along the same
line. The three reflected rays appear to come from a point behind the mirror,
which locates the tip of the virtual image. Convex mirrors are used for
surveillance in stores, where they can provide wide-angle view of the interior.
The mirror on the passenger side of many cars that reads "Caution: objects in
mirror are closer than they appear" is also a convex wide-angle view mirror.
R1
R3
Figure 5.20 - Image
formed by a convex mirror.
R2
image
object
focal point
center of
curvature
We can predict image location if the object distance and mirror focal
length are known. It can be shown that the mirror equation has the same form as
the thin lens equation
(5.9)
1 1 1
+ =
do f di
Magnification is also calculated using the same equation familiar from lens
problems
M=
104
di
do
Lenses and Mirrors
It may surprise you to learn that the mirror equation also applies to a
plane mirror. Since the radius of curvature of a flat surface is infinite, the focal
length of the mirror is also infinite (see Equation 5.8). This means that the 1/f
term in the mirror equation is zero. The resulting equation says that object
distance equals image distance and magnification is one, as you know from your
experience with common household mirrors.
EXAMPLE 5.8
An object is placed 25 cm in front of a concave spherical mirror with a 15
cm focal length. Find the location and size of the image.
Solution
The object distance is negative, with light going left to right from the object
toward the mirror. We expect a real, inverted image since the object is
beyond the focal point.
a. Image distance
1
1
1
+ =
do
f
di
1
1
1
+
=
!25cm 15cm di
di = 37.5cm
15 cm
25 cm
The image forms 37.5 cm to the left of the mirror (on the same side as the
object). Since the light forming the image goes from right to left, positive
image distances locate images to the left of the mirror. This is a real image,
as expected, that can be projected onto a screen.
b. Magnification
M=
M=
hi di
=
ho do
di 37.5cm
=
= !1.5
do !25cm
The image is 1.5 times the size of the object. The negative sign means that
the image is inverted (upside down) compared to the object, as expected.
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LIGHT: Introduction to Optics and Photonics
REFERENCES
For derivations of the thin lens equation and detailed information on aberrations
and thick lens equations see
1. Pedrotti, F., Pedrotti, L.M., Pedrotti, L.S. Introduction to Optics Ed. 3,
Upper Saddle River, NJ: Prentice Hall, 2006.
2. Meyer-Arendt, J. Intro to Classical and Modern Optics. Upper Saddle
3. Hecht, E. Optics, Ed. 4. San Francisco: Addison-Wesley, 2001.
WEB SITES
1. For an explanation of the optics of an eyeglass prescription, see
www.sightandhearing.org/news/healthissue/archive/hi_0303.asp
2. For an illustration of ray tracing, an excellent Java applet is found at
www.hazelwood.k12.mo.us/~grichert/optics/intro.html
3. Lens design and optical CAD software are important tools of optical
engineers. You can find information and some demonstrations and/or
tutorials at any of these web sites.
www.photonengr.com/
www.breault.com/
www.opticalres.com/
www.lambdares.com
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Lenses and Mirrors
REVIEW QUESTIONS AND PROBLEMS
QUESTIONS
Lenses/Thin lens equation
1. What is the difference between a real and virtual focus? Real and virtual image?
2. How does spherical aberration cause blurring of images?
3. What causes chromatic aberration? How does it affect an image?
4. Is it possible to make a real image with a single diverging lens? Explain.
5. For this problem you will need to draw a ray diagram to scale. If an object is placed
10 cm in front of a lens that has a 5 cm focal length, where does the image form? If
the object is moved closer to the focal point, does the image move in toward the lens
or away from the lens?
6. A child gets her first pair of eyeglasses and tries to set a small pile of leaves on fire
by focusing sunlight. Try as she may, the light won't focus. What do you think the
problem might be?
Mirrors
7. A concave mirror has a focal length of 20 cm. What type of image (real/virtual,
upright/inverted) is formed if the object is a) 30 cm in front of the mirror? b)10 cm in
front of the mirror?
8. Can a convex mirror ever form a real image? Why or why not?
9. Why are objects in the passenger side mirror closer than they appear?
LEVEL I PROBLEMS
Lenses/Thin lens equation
10. The thin lens at right has a radius of curvature on the left side of +20 cm and on the
right side of -30 cm. It is made of glass with an index of refraction of 1.58.
a.
What is the focal length of the lens when it is used in air?
b.
Suppose the radii are switched—that is, the left side is -30 cm and the
right side is +20 cm. What would the focal length be? What kind of
lens would it be? -
11. A biconvex lens made of an unknown material has a 35 mm focal length. The radius
of curvature of the first surface is 30 mm and the radius of curvature of the second
surface is -50 mm. Find the index of refraction of the lens material. Draw a diagram
that represents the lens.
12. An object 1 cm high is placed 30 cm in front of a thin lens with a focal length of 15
cm, as shown at left. Where is the image and what is its magnification? Is the image
real or virtual? Is the image erect or inverted?
107
object distance
focal point
15 cm
LIGHT: Introduction to Optics and Photonics
13. Using the same lens as in problem #12 (15 cm focal length) calculate what happens
to the image as the object is placed 20 cm, 10 cm and 5 cm in front of the lens. What
happens to the image location and image type (real or virtual) as the object is moved
from 20 cm toward the lens? What happens to the image orientation and size as the
object is moved from 20 cm toward the lens?
14. Repeat Problem #13 for the same object placed in front of a diverging lens of focal
length -15 cm. Find the image distance, magnification and type of image at distances
of 30, 20, 10 and 5 cm from the lens. What happens to the image location and image
type (real or virtual) as the object is moved from 20 cm toward the lens? What
happens to the image orientation and size as the object is moved from 20 cm toward
the lens
15. A real object 2 cm tall is located 55 cm from a lens of focal length of 7.5 cm. Find
the image distance and determine whether the image is real or virtual. What is the
height of the image? Is the image inverted or upright?
16. A real object 3 cm tall is located 7.5 m from a converging lens of focal length 4.5
cm. Find: image distance, real or virtual object, magnification, height of image,
inverted or upright.
17. A 1.7 m tall person is standing 2.5 m in front of a camera. The camera uses a
converging lens with a focal length of 0.0500 m. Find the image distance. Is the
image real or virtual? What is the magnification and height of the image on the film?
Is the image inverted or upright?
Mirrors
18. A man is shaving with his face 25 cm in front of a concave mirror. The image is 65
cm behind the mirror. Find the focal length of the mirror and its magnification.
19. A dentist's mirror is placed 1.5 cm from a tooth. The enlarged image is located 4.3
cm behind the mirror. What kind of mirror (plane, convex, concave) is being used?
Determine the focal length of the mirror, magnification and image orientation
relative to the object.
20. A grocery store uses convex mirrors to monitor the aisles in a store. The mirrors
have a radius of curvature of 4 m. What is the image distance if a customer is 15 m
in front of the mirror? Is the image real or virtual? If a customer is 1.6 m tall, how
tall is the image? LEVEL II PROBLEMS
Lenses/thin lens equation
21. What is the object distance and object height if the magnification is 6 and the real
image is 30 cm from the lens and 1 cm high?
22. Two positive lenses, each of focal length f = 3 cm, are separated by a distance of 12
cm. An object 2 cm high is located 6 cm to the left of the first lens. Find the final
image characteristics, distance, height and orientation. Draw a diagram.
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Lenses and Mirrors
23. A converging lens with f =12 cm is 28 cm to the left of a diverging lens f = -14 cm.
An object is located 6 cm to the left of the converging lens. Find the final image
distance and the overall magnification.
24. Two 20 cm focal length lenses are placed on an optical bench 70 cm apart. A 1 cm
high object is placed 40 cm to the left of the first lens. Find the location of the
images of the first lens and the second lens. Find the overall magnification of the two
lens system
25. Repeat problem #24 with the distance between the lenses reduced to 50 cm.
26. Repeat problem #24 with the distance between the lenses reduced to 30 cm.
27. A converging lens projects a real image on a screen, which can serve as the object
for a second lens. Show how this method can be used to determine the focal length of
a negative (diverging) lens.
109