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2 Reflection From A Plane Mirror: The Law Of Reflection

2 Reflection From A Plane Mirror: The Law Of Reflection

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Figure 24.4



A Rays reflecting

from a flat surface. A surface can

be considered flat if its roughness

is smaller than l. This mirror is

called a plane mirror. B The angle

of reflection is equal to the angle of

incidence.



Normal

direction

Incident

ray



Reflection from a flat surface



ui ur



Figure 24.4A describes the case of an incoming plane wave, so the rays are parallel and strike the surface at many different points. We can characterize this reflection by considering only a single ray as shown in Figure 24.4B. (For a plane wave,

the other rays are parallel as in Fig. 24.4A and reflect at the same angle.) The vertical dashed line in Figure 24.4B is drawn perpendicular (also called “normal”) to

the mirror, and we can measure the directions of the incoming and outgoing rays

relative to this perpendicular direction. The incoming ray is called the incident

ray, and the angle it makes with the surface normal is the angle of incidence ui. The

outgoing ray is called the reflected ray, and its angle with the normal is the angle

of reflection ur . For reflection from a flat surface, the angle of incidence is equal to

the angle of reflection,



Rough surface



The reflected rays will not all be

parallel to one another. Compare

with Figure 24.4.



ui ϭ ur



Figure 24.5 Rays are reflected

in different directions by different

parts of a rough surface, forming a

diffuse reflection.



Extrapolate

rays back

to apparent

point of

origin.



ur

ui



i



Object

L



Image

L



Figure 24.6 Forming an image

with a plane mirror. Rays from

the object (the arrow on the left)

reflect from the mirror and reach

your eye. These rays appear to

have come from the image, on the

right of the mirror. It is a virtual

image because light does not actually pass through the image point.

This sketch shows only rays from

the top and bottom of the object,

but rays from other points on the

object also contribute to the image.

798



(24.1)



This equation, called the law of reflection, can be derived from our general principle concerning the reversibility of the propagation of light. Suppose the angle of

reflection is smaller than ui. If we were to reverse the outgoing ray, its angle of reflection would be even smaller and the new reflected ray would not retrace the original

incoming ray. The only way the reflection process can be reversible is for the angle

of reflection to equal the angle of incidence (Eq. 24.1).

Reflections from a perfectly flat mirror (Fig. 24.4) are called specular reflections. If the reflecting surface is rough, we must consider reflections from all the

individual pieces of the surface. In that case, an incident plane wave will give rise to

many reflected rays propagating outward in many different directions (Fig. 24.5).

This reflection is called a diffuse reflection.



Image Formation by a Plane Mirror



ui ϭ ur

for each ray



o



ui ϭ ur



B



A



Observer’s eye



Reflected

ray



When you view an object through its reflection in a mirror, you are viewing an

image of the object. An image formed by a plane mirror is shown in Figure 24.6.

Here we show a particularly simple object (an arrow), but the object could be a tree

or a person. We will usually use an arrow as our example object because we’ll often

be interested in the orientation of the image relative to the object. As in Figure 24.3,

an infi nite number light rays emanate from each point on the object, although we

only show two rays emanating from the top and bottom here. Some of these rays

strike the mirror and are reflected so as to reach your eye, which then uses the rays

to form an image. To your eye, the location of this image—that is, the location of

the arrow—is the point from which these rays appear to emanate. This point can

be found by extrapolating the rays back in a straight-line fashion to their apparent

point of origin as shown by the dashed lines in Figure 24.6. This is our fi rst example

of the ray-tracing procedure. Since each ray obeys the law of reflection (Eq. 24.1),

we can use that together with some geometry to determine the image location. A

few rays are shown in Figure 24.6 for an object located a distance L in front of the

mirror. When the rays that reach the observer’s eye are extrapolated back to their

apparent common point of origin, the image is found to be a distance L behind the

mirror.

In addition to locating the image, Figure 24.6 gives several other results. First,

the size of an image formed by a plane mirror is equal to the size of the object; that

is, the height of the image hi is equal to the height of the object ho. Second, because



CHAPTER 24 | GEOMETRICAL OPTICS



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the image point is located behind the mirror, light does not actually pass through

the image. For this reason, the image is called a virtual image.



E X A M P L E 2 4 .1



Reflection from a Corner Cube



Definition of a virtual image



Insight 24.1



Figure 24.7A shows a combination of two plane mirrors connected to form a right

angle. In three dimensions, one can form a similar structure called a corner cube by

combining mirrors to make three faces of a cube. The light ray in Figure 24.7A is incident at an angle ui relative to the direction normal (perpendicular) to mirror 1. Find

the angle of the outgoing light ray uout relative to this normal direction.

RECOGNIZE T HE PRINCIPLE



To analyze reflection from a single plane mirror, we must fi rst determine the angle of

incidence, the angle the incident ray makes with the direction normal to the mirror.

According to the law of reflection, the angle of reflection is then equal to the angle of

incidence. Here we must use this procedure fi rst for the reflection from mirror 1 and

then a second time for the reflection from mirror 2.



WHAT DETERMINES IF A SURFACE

WILL REFLECT LIGHT?

The way a surface does or does not

reflect light depends on the material

making up the surface. The electric

and magnetic fields of the light wave

exert forces on electrons and ions near

the surface. A typical mirror consists

of a metal surface, and the motion

of electrons and ions induced by the

electric and magnetic fields produces

the reflected light wave.



SK E TCH T HE PROBLEM



Figure 24.7A shows the incident ray along with the normal direction for mirror 1

(shown as the dashed line). Reflection from mirror 1 gives an outgoing ray making an

angle ur with the normal to mirror 1 as sketched in Figure 24.7B. This ray is then the

incident ray for reflection from the bottom mirror. We must apply the law of reflection

(Eq. 24.1) for each reflection in Figure 24.7B.

IDENT IF Y T HE REL AT IONSHIPS



The angle ui in Figure 24.7A is the angle the incoming ray makes with the normal

to its reflecting surface, so it is also the angle of incidence in the law of reflection,

Equation 24.1. The angle of reflection from this fi rst surface is thus ur ϭ ui. The two

reflecting surfaces (the two mirrors) are at right angles, so the dashed lines showing

the directions normal to each mirror in Figure 24.7B also form a right angle. Because

the interior angles of a triangle add up to 180°, the angle of incidence for the second

reflection is



Mirror 1



ui



ui2 ϭ 90° Ϫ ur ϭ 90° Ϫ ui

Applying the law of reflection to the ray as it reflects from mirror 2, we fi nd



ur2 ϭ ui2 ϭ 90° Ϫ ui



Mirror 2



(1)

A



SOLV E



From the geometry in Figure 24.7B, the angle the outgoing ray makes with the horizontal is



Mirror 1



uout ϭ 90° Ϫ ur2



ui



Combining this with Equation (1) gives



ur



uout ϭ 90° Ϫ ur2 ϭ 90° Ϫ (90° Ϫ ui)

uout ϭ



Incident

ray



ur



uout

uout

ur2

Mirror 2

ui2 ϭ 90Њ Ϫ ur



ui



The outgoing ray is thus parallel to the incident ray.



What does it mean?

An interesting feature of this result is that the outgoing ray is always parallel to

the incident ray for any value of ui. A combination of mirrors called a corner cube,

the three-dimensional version of Figure 24.7, has a similar property: its reflected

rays are always parallel to the incident rays. Such devices, called retroreflectors,



B



Figure 24.7 Example 24.1.

Reflection from two mirrors that

form a right angle. A The incident

ray. B Reflected rays and the corresponding angles of incidence and

reflection.



24.2 | REFLECTION FROM A PLANE MIRROR: THE LAW OF REFLECTION



799



© Paul Ridsdale/Alamy



are useful in many applications. For example, the Apollo astronauts who visited

the Moon left a number of corner cube reflectors on its surface. When laser light

from the Earth strikes one of these corner cubes, the reflected light travels in a

path parallel to the incident ray and back to the Earth. The time delay between

incident and reflected light pulses can be used to measure the Earth–Moon distance with high precision. (See Problem 23.61, page 794.) In principle, a single,

flat mirror would reflect light in the same way, but the mirror would have to be

aligned very precisely; otherwise, the reflected light would miss the Earth. With

a retroreflector, the reflected light is always parallel to the incident ray, no matter

what the alignment. A more down-to-Earth application of retroreflectors is found

on many roads. Most roadside reflectors, and even some paints, contain tiny

crystals that act as corner cubes to reflect the rays from your headlights back to

you, thus giving a noticeable indication of the edges of the road. That is also how

reflective clothing works (Fig. 24.8).

Figure 24.8 Reflective clothing

contains tiny corner cubes, each of

which reflects light back parallel

to the incident rays as in Figure

24.7B. This cyclist’s vest seems

to “glow” in the dark because it

reflects light (such as from a car’s

headlights) back toward its source.



CO N C E P T C H E C K 2 4 .1 | Reflection from Two Mirrors

Consider two plane mirrors that make an angle of 60° with each other (Fig. 24.9).

An incident ray is parallel to the bottom mirror. This ray reflects from the mirror

at the left and then strikes the bottom mirror. What angle of incidence ui2 does

the reflected ray make with the bottom mirror, (a) 30°, (b) 45°, or (c) 60°?



24.3

ui1

Parallel

60Њ



ui2 ϭ ?



Figure 24.9 Concept Check

24.1.



Incident ray



Reflected ray

u1 ur



Air

Glass



ur ϭ u1



u2

Refracted ray



Figure 24.10 An incident light

ray is both reflected and refracted

when it strikes the surface of a

transparent object such as a piece

of glass. A portion of the light

energy is reflected with ur ϭ u1,

whereas the rest is refracted as it

passes into the glass.



800



|



REFR AC TION



When a light ray strikes a transparent material like a windowpane, some of the

light is reflected (according to the law of reflection described in Section 24.2), and

the rest passes into the material. Figure 24.10 shows a light ray incident on a piece

of glass with two flat, parallel sides. The ray that travels into the glass is called a

refracted ray.



The Index of Refraction and Snell’s Law

The direction of the refracted ray is measured using the angle u2 this ray makes with

the surface normal. The value of u2 depends on the incident angle (which we now

denote as u1) and also on the speed of light within the material. The speed of light

in vacuum is c ϭ 3.00 ϫ 108 m/s. When light travels through a material substance,

however, the associated electric and magnetic fields interact with the atoms of the

substance, and this interaction affects the speed of the wave. As a result, the speed

of light inside a substance such as glass, water, or air is less than the speed of light

in vacuum. Table 24.1 lists the speed of light in several substances.

Figure 24.11A shows a plane wave initially traveling in a vacuum and then striking the flat surface of a piece of glass; this sketch illustrates how the change in the

speed of light in going from vacuum to glass affects the direction of the rays inside

the glass. Here we show several rays along with some associated wave fronts in

both the vacuum and the glass. The incoming rays make an angle u1 relative to the

surface normal, and the incident wave fronts are perpendicular to these rays. The

incident light travels at speed c, the speed of light in a vacuum. The light inside the

glass, however, moves at speed v, which is less than c.

From Section 24.1 (Fig. 24.2), the perpendicular distance between wave fronts

is proportional to the wave speed. The wave fronts in Figure 24.11A are drawn at

successive, equally spaced moments in time, so the spacing between wave fronts is

ct on the vacuum side and vt in the glass. Figure 24.11B shows an expanded view

of two adjacent wave fronts, and two right triangles are indicated. On the vacuum

side, the right triangle shaded in red has sides L and ct, with angle u1 adjacent to

side L. From the geometry of this triangle, we have



CHAPTER 24 | GEOMETRICAL OPTICS



Copyright 2009 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.



Speed of Light and Index of Refraction for Several Substances

for Room Temperature and Yellow Light (L ‫ ؍‬550 nm)



Ta b l e 2 4 . 1



Substance



Speed of Light (m/s)



Vacuum



2.998 ϫ



Air



2.997 ϫ 108



Index of Refraction



108



1.0000 (exactly)

1.0003



Water (liquid)



2.26 ϫ 108



1.33



Water (ice)



2.29 ϫ



108



1.31



Benzene



2.00 ϫ



108



1.50



Quartz crystal



2.05 ϫ 108



1.46



Diamond



1.24 ϫ 108



2.42



Pyrex glass



2.04 ϫ



108



1.47



Plexiglas (plastic)



2.01 ϫ



108



1.49



vitreous humor



2.23 ϫ 108



1.34



aqueous humor



2.26 ϫ



108



1.33



cornea



2.17 ϫ



108



1.38



lens



2.13 ϫ 108



1.41



Regions in the eye:



sin u 1 5



ct

L



(24.2)



There is a corresponding triangle shaded in yellow on the glass side with edge

lengths L and vt and an angle u2 adjacent to side L, which leads to

sin u 2 5



vt

L



(24.3)



We can rearrange each of these relations to solve for L, resulting in L ϭ ct/sin u1 and

L ϭ vt/sin u2 . Setting them equal, we fi nd

ct

vt

5

sin u 1

sin u 2

The factor of t can be canceled, giving

sin u 1 5



Ray 1



Ray 2



Ray 3



u1

ct



nt 1

e fro

Wav

nt 2

e fro

Wav

nt 3

e fro

Wav

Vacuum



Glass



vt



c

sin u 2

v



(24.4)



u1

ct

Vacuum



u1

vt



Glass



u2



Figure 24.11 A Refraction

occurs due to a difference in the

wave speeds in two regions, which

causes the rays to change direction

at the interface. B Derivation of

Snell’s law. The speed of light in a

vacuum is c, whereas the speed of

light in glass is v.



L

u2



u2

A



B



24.3 | REFRACTION



801



Incident ray



Reflected ray

u1 u1



The ratio c/v is called the index of refraction of the glass and is denoted by

n5



n1 Ͻ n2



n1

n2



c

v



(24.5)



We can thus rewrite Equation 24.4 as

u2



sin u 1 5 n sin u 2



Refracted ray

SNELL’S LAW:

n1 sin u1 ϭ n2 sin u2



In Figure 24.11, we assumed there is vacuum on the incident side and glass on the

refracted side. A common situation involves light passing between two substances

such as air and glass or air and water. If the two substances have indices of refraction n1 and n2 (Fig. 24.12), the incident and refracted angles are related by

n1 sin u 1 5 n2 sin u 2



Figure 24.12 Snell’s law gives

the relationship between the angle

of incidence and the angle of

refraction.



(24.6)



(24.7)



This relation between the incident and refracted angles is called Snell’s law. Table

24.1 lists the indices of refraction for some common substances. Since n is the ratio

of two speeds (Eq. 24.5), the index of refraction is a dimensionless number.



EXAMPLE 24.2



Refraction on a Pier



A lobster fisherman looks just over the edge of a pier and spots a lobster resting at the

bottom. This fishing spot is d ϭ 3.5 m deep. Judging from the angle at which he spots

the lobster, the fisherman thinks the lobster is a horizontal distance L app ϭ 6.0 m from

the shore, but when he drops his trap at that location, he does not catch the lobster.

What is the true horizontal distance of the lobster from the shore?

RECOGNIZE T HE PRINCIPLE

u2

O

u1

u2

d

True

position



Water

A



Apparent

position

of lobster

B



Ltrue

Lapp



The fisherman determines the lobster’s apparent position by

extrapolating the ray that meets his eye back to its apparent

point of origin at the bottom of the harbor. This extrapolation is shown by the dashed line in Figure 24.13. However,

because the ray from the lobster is refracted when it passes

from the water into the air, the actual position of the lobster

is much closer to the shore. We can use Snell’s law to fi nd

the angle u1 that the true ray from the lobster makes with

the surface normal, and from that we can locate the true

position of the lobster.

SK E TCH T HE PROBLEM



Figure 24.13 shows the problem, including the apparent

position of the lobster.



Figure 24.13 Example 24.2.



IDENT IF Y T HE REL AT IONSHIPS

Insight 24.2

REVERSIBILIT Y OF LIGHT

RAYS AND REFRACTION

In Figure 24.12, light is incident at

an angle u1 from side 1, leading to a

refracted angle u2 given by Snell’s law

(Eq. 24.7). If the light had instead

been incident from side 2 (from the

bottom of the figure) at an incident

angle u2 , applying Snell’s law would

give a refracted angle u1. Hence, in

accord with the general principle of

reversibility of propagation direction

stated in Section 24.1, the refraction

of light is reversible.



802



To apply Snell’s law, we denote the angle of incidence (on the water side of the water–

air interface) as u1 and the angle of refraction (on the air side of the interface) as u2 .

The angle u2 is related to L app, the apparent distance of the lobster from the shore, by



tan u 2 5



Lapp

d



because L app and d are two sides of the right triangle OAB in Figure 24.13. Inserting

the given values d ϭ 3.5 m and L app ϭ 6.0 m, we fi nd



u 2 5 tan21 a



6.0 m

b 5 60°

3.5 m



which is also the angle of refraction for the ray that travels from the lobster to the

fisherman. From Snell’s law, we thus have n1 sin u 1 5 n2 sin u 2, where n1 is the index



CHAPTER 24 | GEOMETRICAL OPTICS



of refraction of water and n 2 is the index of refraction for air. Using the values of n1

and n 2 from Table 24.1 along with our value of the angle u2 , we have



sin u 1 5



n2

1.00

sin u 2 5

sin 1 60° 2

n1

1.33

© Cengage Learning/Charles D. Winters



u 1 5 41°

SOLV E



Now that we have the value of u1, we can fi nd the lobster’s true distance from the

shore, L true. Using the right triangle shaded in green in Figure 24.13, we have



tan u 1 5



Ltrue

d



Ltrue 5 d tan u 1 5 1 3.5 m 2 tan 1 41° 2 5 3.0 m

What does it mean?

The lobster is thus much closer to shore than it appears to the naive fisherman,

which would affect where he drops his trap to have a chance of catching the lobster. This effect is illustrated in Figure 24.14, which shows a pencil inserted into a

glass of water. The pencil appears to bend where it enters the water due to refraction of light rays as they leave the water.



Figure 24.14 This pencil is

partially immersed in water. It

appears to “bend” under the

water’s surface due to refraction

of rays that pass from the water

into the air.



Applying Snell’s Law

When a light ray strikes a plane surface, the angle of the reflected ray is given by

the law of reflection (Eq. 24.1) and the angle of the refracted ray is given by Snell’s

law (Eq. 24.7). Figure 24.12 shows the case in which the incident ray comes from

the side with the smaller index of refraction, whereas in Figure 24.15 the light is

incident from the side with the larger index. For instance, substance 1 (the lower

substance in Fig. 24.15) might be water or glass and substance 2 could be air.

Snell’s law reads the same whether light begins in the substance with the larger

or smaller index of refraction,

n1 sin u 1 5 n2 sin u 2



(24.8)



Possible angles of incidence always lie between zero and 90°, so 0° # u 1 # 90°

and 0° # u 2 # 90°. For these angles, the function sin u increases as u is increased.

Hence, according to Equation 24.8, the side with the larger index always has the

smaller angle. In words, we say that light is refracted toward the normal direction

when moving into the substance with the larger index of refraction (Fig. 24.12).

Light is refracted away from the normal direction when moving into the substance

with the smaller index of refraction (Fig. 24.15).

Refracted ray



CO N C E P T C H E C K 2 4 . 2 | Snell’s Law

Light travels from a vacuum (n1 ϭ 1.00) into a plate of glass with n 2 ϭ 1.55,

with an angle of incidence of 30°. Is the angle of refraction (a) larger than 30° or

(b) smaller than 30°? Also, use Snell’s law to calculate the angle of refraction.



Total Internal Reflection

Let’s now consider light incident from the side with the larger index of refraction

in Figure 24.15 (the bottom side) a little more carefully. Because the index n1 is

greater than n 2 , the angle of refraction u2 is greater than u1 and light is refracted

away from the normal direction. As the incident angle u1 is made larger and larger,

the refracted angle u2 increases more and more. Eventually, u2 will reach 90° and the

refracted ray will emerge parallel to the surface. The value of the incident angle at

which that occurs is called the critical angle. If the incident angle u1 in Figure 24.15



u2



n2

n1

u1

Incident ray



n1 Ͼ n2



u1

Reflected ray



Figure 24.15 Reflection and

refraction when light travels from

a region of higher to a region of

lower index of refraction. Here

n1 Ͼ n 2; region 1 at the bottom

might be glass, and region 2 at the

top might be air. Compare with

Figure 24.12, in which n1 Ͻ n 2 .

24.3 | REFRACTION



803



Figure 24.16 An example of



Courtesy of The Harvard Science Center. ©

The President & Fellows of Harvard College.



total internal reflection. Light from

a laser at the lower left is reflected

at the air–water interface. There is

no refracted ray.



is increased beyond the critical angle, Snell’s law has no solution for the refracted

angle u2 . Physically, there is no refracted ray, and all the energy from the incident

light ray is reflected. This behavior, called total internal reflection, is possible only

when light is incident from the side with the larger index of refraction.

Figure 24.16 shows an example of total internal reflection. Here light from a

laser travels into the side of a tank of water. The light is reflected at the interface

between the water and the air, but the angle of incidence exceeds the critical angle,

so there is no refracted ray. Hence, no light gets through this otherwise “transparent” interface!

We can fi nd the value of the critical angle using Snell’s law. When the angle of

incidence equals the critical angle, the angle of refraction in Figure 24.15 is u2 ϭ 90°

(because the direction of the refracted ray would be parallel to the interface). Inserting this angle into Snell’s law (Eq. 24.8) and using sin(90°) ϭ 1, we get



Total internal reflection



n1 sin u 1 5 n2 sin u 2



n1 sin u 1 5 n2 sin u 2 5 n2 sin 1 90o 2 5 n2

sin u 1 5



n2

n1



Denoting the critical angle as ucrit (ϭ u1) gives

sin u crit 5



n2

n1



or

u crit 5 sin 21 a



Total internal reflection



Light is confined to an optical

fiber by total internal reflection.

A



Figure 24.17



A Total internal

reflection of light traveling inside

an optical fiber. B Light escapes

from an optical fiber only through

the end.



804



© Tetra Images/Jupiterimages



Optical

fiber



B



CHAPTER 24 | GEOMETRICAL OPTICS



n2

b

n1



(24.9)



The value of ucrit thus depends on the ratio of the indices of

refraction of the substances on the two sides of the interface.

When the angle of incidence is equal to or greater than the

critical angle, light is reflected completely at the interface.

Amazingly, this phenomenon can occur even when the interface between the two substances would seem to be completely

transparent (such as between water and air). Look for a reflection from such an interface the next time you go swimming or

gaze through a window.

Total internal reflection is used in fiber optics. Optical fibers

are composed of specially made glass and are used to carry

telephone and computer communications signals. These signals are sent as light waves, using total internal reflection as

sketched in Figure 24.17 to keep the light from “leaking” out

the sides of the fiber. Optical fibers are designed to carry sig-



© Stephen St. John/National Geographic/Getty Images



nals over very long distances, so to minimize signal loss it is important that these

reflections be as perfect as possible. (The model of an optical fiber shown here is

highly simplified; we’ll describe a more realistic picture in Chapter 26.) A mirage is

another example of total internal reflection (Fig. 24.18). A mirage is formed by total

reflection of light between layers of cool air and warm air.

CO N C E P T C H E C K 2 4 . 3 | When Is Total Internal Reflection Possible?

In which of the following cases is it possible to have total internal reflection?

(a) Light traveling from water into air

(b) Light traveling from air into water

(c) Light traveling from glass into water

(d) Light traveling from water into diamond

(e) Light traveling from benzene into quartz (Hint: See Table 24.1.)



EXAMPLE 24.3



Total Internal Reflection

between Water and Air



Figure 24.18 A mirage forms

when there is a layer of warm air

next to the ground. Light from this

car undergoes total internal reflection from this layer.



Find the critical angle for total internal reflection between water and air.

RECOGNIZE T HE PRINCIPLE



When we used Snell’s law to derive Equation 24.9, we saw how to calculate the critical angle. The purpose of this problem is to get a feel for typical values of the critical

angle.

SK E TCH T HE PROBLEM



Figure 24.19A shows light incident on the interface between water and air. The

incident ray is at the critical angle, so the refracted ray (if it existed) would be

parallel to the interface and u2 ϭ 90°.

IDENT IF Y T HE REL AT IONSHIPS



We can fi nd the critical angle using the result for ucrit in Equation 24.9:



u crit 5 sin 21 a



nair

n2

b 5 sin 21 a

b

n1

nwater



Incident

ray



u1 = ucrit



Reflected

ray



A



u1 > ucrit



SOLV E



When u1 > ucrit, the interface acts

as a perfect mirror.



Inserting the values nair ϭ 1.0003 Ϸ 1.00, and nwater ϭ 1.33, we get



nair

1.00

b 5 sin 21 a

b 5 sin 21 1 0.75 2 5

nwater

1.33



Refracted

ray



Water



We can get values for the indices of refraction for water (nwater) and air (nair)

from Table 24.1.



u crit 5 sin 21 a



u2



Air



49°



B



What does it mean?

The incident ray in Figure 24.19A is drawn at ucrit ϭ 49°, showing that the incident ray does not have to be “close” to the interface to achieve total internal

reflection. You can test this result the next time you go swimming. If you are

underwater and look in a direction for which the angle of incidence is greater than

the critical angle (Fig. 24.19B), the water–air interface will appear as a perfect

mirror. The swimmer in Figure 24.19B will see the fish reflected in this interface

mirror. Try it.



Figure 24.19 Example 24.3.

A Total internal refl ection at an

air–water interface. B When a

swimmer (on the right) looks at

light coming from the air–water

interface, all light with an angle of

incidence greater than the critical

angle is reflected. There are no

rays incident from the air that give

rays at this angle; the interface

thus acts as a perfect mirror when

viewed at this angle.



24.3 | REFRACTION



805



Blue



Yellow Red



First refraction



v (m/s)

1.95 ϫ 108



Crown glass



Incident light

(red ϩ blue)



1.52

1.50



Plexiglass



Second

refraction



© David Parker/Photo

Researchers, Inc.



n

1.54



Red

Blue



2.00 ϫ 108

A



B



1.48

Fused quartz

1.46



400



500 600

l (nm)



Figure 24.21 A The angle of refraction depends on the wavelength (i.e., the

color) of the light. Blue light and red light incident from the left will therefore

refract at different angles and emerge from the prism propagating in different

directions. B This enables a prism to separate different colors in an incident

beam of white light.



2.05 ϫ 108



700



Figure 24.20 Variation of the

speed of light (right axis) and the

index of refraction (left axis) as a

function of wavelength for several

materials.



Dispersion

The direction of a refracted ray depends on the indices of refraction of the materials

on the two sides of an interface, and these indices of refraction depend on the speed

of light in each material (Eq. 24.5). The speed of light in a vacuum has the constant

value c for all wavelengths, that is, for all colors. When light travels in a material,

however, the speed depends on the color of the light. This dependence of wave speed

on color is called dispersion, and the variation of the speed of light with wavelength

for several materials is shown in Figure 24.20.

For example, the index of refraction for red light in quartz is nred ϭ 1.46 and that

for blue light is nblue ϭ 1.47. The difference nblue Ϫ nred is much smaller than the difference between the (average) index of refraction for quartz and water (Table 24.1).

This difference, however, does mean that the angle of refraction is slightly different

for different colors in quartz (and in other materials, too). This effect is used by a

prism to separate a beam of light into its component colors.

Prisms are typically composed of glass, with a triangular cross section (Fig. 24.21).

In Figure 24.21A, a beam of light is incident on one surface of a prism; we suppose

this light is composed of two colors (red and blue), and we use two rays with different colors to denote the incident light. Each ray is refracted twice: once when it

enters the prism and again when it exits. The fi rst refraction occurs at the left-hand

face of the prism. Because the refractive index of glass depends on the color of the

light, incident beams of different color have slightly different angles of refraction.

Blue light has a larger index of refraction (Fig. 24.20) than red light, so the blue ray

inside the glass makes a smaller angle with the normal direction on the left than

does the red ray. Hence, inside the glass the red and blue components travel in different directions. This difference in propagation direction is increased at the second

refraction, when the light leaves the prism. The calculation of these outgoing angles

involves two applications of Snell’s law, one for each surface of the prism. (We leave

this calculation for the end-of-chapter problems; see Problems 29–31.) The result is

shown in Figure 24.21B. In words, we say that the prism has “dispersed” the light

into different directions according to color. That is why the variation of the index

of refraction with wavelength (Fig. 24.20) is called dispersion.



CO N C E P T C H E C K 2 4 . 4 | Refraction of Rays with Different Colors

A mixture of red light and blue light is incident from vacuum onto a thick plexiglass slab. If nred ϭ 1.49 and nblue ϭ 1.51, which color will have the larger angle

of refraction, (a) the red light or (b) the blue light? Or, will they (c) have the same

angle of refraction? Hint: You should be able to answer this question without

working out values of the angles of refraction.



806



CHAPTER 24 | GEOMETRICAL OPTICS



EXAMPLE 24.4



Red ϩ blue



Dispersing Light by Refraction



Figure 24.22 shows light incident from air onto a flat quartz slab. This light is a mixture of red and blue, denoted in the figure by the two incident rays with these colors.

Both rays have an angle of incidence u1 ϭ 45.0°. Find the angles of refraction for the

two rays in the quartz plate. Assume the indices of refraction for red and blue light

in quartz are n quartz, red ϭ n 2, red ϭ 1.459 and n quartz, blue ϭ n 2, blue ϭ 1.467. Give your

answers to three significant figures.

RECOGNIZE T HE PRINCIPLE



n1



u1



Air

Direction

of incident

rays



Quartz

n2, blue n2, red

u2, blue

u2, red



Figure 24.22 Example 24.4.



We can fi nd the angle of refraction for each color using Snell’s law and the values of

the index of refraction for each color. The point of this example is to get a feeling

for how much the two rays are “dispersed.” How large is the difference between the

directions of the two rays?

SK E TCH T HE PROBLEM



Figure 24.22 shows the problem. The index of refraction in quartz is greater for blue

light than for red light, so the blue ray is refracted closer to the normal as shown in

the figure.

IDENT IF Y T HE REL AT IONSHIPS



The indices of refraction for red and blue light in quartz are given. In air, the indices of refraction for both colors are the same to four significant figures (Table 24.1):

nair, red ϭ nair, blue ϭ n1 ϭ 1.000.

SOLV E



We can fi nd the angles of refraction using Snell’s law (Eq. 24.7). For the red light ray,

we have



n1 sin u 1 5 n2, red sin u 2, red

Solving for u2, red, we get



sin u 2, red 5



n1

n2, red



sin u 1



(1)



Inserting the values of the indices of refraction and u1 gives



sin u 2, red 5



n1

1.000

sin 1 45.0° 2

sin u 1 5

n2, red

1.459

u2, red ϭ 29.0°



(2)



We have kept three significant figures because, as we’ll see below, the difference

between the refracted angles for the two rays will be small.

We compute the angle of refraction for the blue light ray in the same way. Equation

(1) leads to



sin u 2, blue 5



n1

1.000

sin u 1 5

sin 1 45° 2

n2, blue

1.467

u2, blue ϭ 28.8°



(3)



What does it mean?

The angles of refraction of the two rays differ by only a small amount, just 0.2°,

but that is enough to give the dispersion of light by a prism in Figure 24.21B. (To

fi nd the angle of the outgoing ray for a prism, we must also include the refraction



24.3 | REFRACTION



807



of the light rays at the second surface of the prism.) Also, the angle of refraction

is smaller for blue light (see Concept Check 24.4), so the blue ray is closer to the

normal direction. Thus, refraction causes blue light to be deflected more (with

respect to the incoming direction) than red light.



24.4



Spherical

mirror

A

Principal

axis

Incoming

rays

Focal

point



F



C



R



f



B

Principal

axis

Focal

point

Center of

curvature



C



F



C



Figure 24.23 A A spherical

mirror matches the shape of a

sphere. The radius R of this sphere

is called the radius of curvature of

the mirror. B If the incoming rays

are parallel to the principal axis

of the mirror, all the reflected rays

converge at the focal point of the

mirror. C If a light source is placed

at the focal point, the outgoing

rays reflected from the mirror are

parallel to the axis.



808



REFLECTIONS AND IM AGES PRODUCED

BY CURVED MIRRORS



In Section 24.2, we saw that a plane mirror can produce an image of an object

(Fig. 24.6). That image and all others formed by plane mirrors are the same size as

the original object. A curved mirror, however, can achieve a magnified image, an

image that appears larger or smaller than the original object. Magnified images are

used in many applications, ranging from telescopes to a car’s rearview mirror, so

let’s now consider how to produce one.



R



Center of

curvature



|



Ray Tracing for Spherical Concave Mirrors

Consider the spherical mirror in Figure 24.23A in which the surface of the mirror

forms a section of a spherical shell (like a beach ball). The radius R of this sphere

is called the radius of curvature of the mirror. The mirror’s principal axis (parts B

and C of Fig. 24.23) is the line that extends from the center of curvature C to the

center of the mirror. The mirror shown in Figure 24.23 is concave, curving toward

objects placed in front of it.

Concave spherical mirrors have two important properties. First, consider Figure

24.23B, which shows incoming light rays directed parallel to the principal axis of

the mirror. If these rays are close to the principal axis, then after reflecting at the

surface of the mirror they all pass through the single point F indicated in Figure

24.23B. (We’ll discuss the behavior of rays far from the axis in Section 24.8.) This

point F, called the focal point of the mirror, is located a certain distance f from the

mirror surface along the principal axis. The distance f is called the focal length of

the mirror. Second, applying the general principle that the propagation of light is

reversible, rays that originate at the focal point reflect from the mirror and propagate outward, parallel to the principal axis as in Figure 24.23C. The mirror thus

“works” both ways: either focusing parallel light rays at the focal point F or generating a set of parallel rays from light that originates at point F and strikes the mirror.

When an object is placed in front of a spherical mirror, light from the object is

reflected from the mirror and forms an image. This is illustrated in Figure 24.24,

which shows photos of objects placed in front of a concave mirror. Accompanying

each photo is a schematic showing the object, the mirror, and the “observer” (the

camera that took the photo). In Figure 24.24A, the object—a candle—is placed

fairly close to the mirror. The observer can see both the original candle and its

image formed by the mirror. Here the image appears larger than the true candle. In

Figure 24.24B, the object—a person’s head—is somewhat farther from the mirror.

The observer now sees the back of the person’s head along with an image of the

person’s face. Note that the face is upside down; we call this an inverted image.

Figure 24.24 shows two examples of images formed by curved mirrors. Let’s now

apply ray tracing to analyze some of the image properties.

Figure 24.25 is a ray diagram for an object placed in front of a curved mirror;

here the object is the upward-pointing arrow. The figure shows rays that emanate

from the tip of the arrow on the left. Although these rays reflect from different

parts of the mirror, they all intersect at a single point, which is the tip of the arrow’s

image. A similar result is found for rays from other points on the object.

We can use the following ray-tracing procedure to find the location of the image

formed by a spherical mirror. Figure 24.26 shows an object far from the mirror,



CHAPTER 24 | GEOMETRICAL OPTICS



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