The Nature of Light and the Laws of Geometric Optics The Nature of Light and the Laws of Geometric Optics Q35.4 With a vertical shop window, streetlights and his own reflection can

35

CHAPTER OUTLINE

35.1 The Nature of Light35.2 Measurements of the Speed of Light35.3 The Ray Approximation in Geometric Optics35.4 Reflection35.5 Refraction35.6 Huygen’s Principle35.7 Dispersion and Prisms35.8 Total Internal Reflection35.9 Fermat‘s Principle

The Nature of Light and theLaws of Geometric Optics

ANSWERS TO QUESTIONS

Q35.1 The ray approximation, predicting sharp shadows, is valid forλ << d . For λ~ d diffraction effects become important, and thelight waves will spread out noticeably beyond the slit.

Q35.2 Light travels through a vacuum at a speed of 300 000 km persecond. Thus, an image we see from a distant star or galaxymust have been generated some time ago. For example, the starAltair is 16 light-years away; if we look at an image of Altairtoday, we know only what was happening 16 years ago. Thismay not initially seem significant, but astronomers who look atother galaxies can gain an idea of what galaxies looked likewhen they were significantly younger. Thus, it actually makessense to speak of “looking backward in time.”

Q35.3 Sun

Moon

Note: Figure not at all to scale

no eclipse

partial eclipse

Earth’s surface total eclipse(full shadow)

no eclipse

FIG. Q35.3

315

316 The Nature of Light and the Laws of Geometric Optics

Q35.4 With a vertical shop window, streetlights and his ownreflection can impede the window shopper’s clear view ofthe display. The tilted shop window can put thesereflections out of the way. Windows of airport controltowers are also tilted like this, as are automobilewindshields.

FIG. Q35.4

Q35.5 We assume that you and the child arealways standing close together. For a flatwall to make an echo of a sound that youmake, you must be standing along a normalto the wall. You must be on the order of100 m away, to make the transit timesufficiently long that you can hear the echoseparately from the original sound. Yoursound must be loud enough so that you canhear it even at this considerable range. Inthe picture, the dashed rectangle representsan area in which you can be standing. Thearrows represent rays of sound.

Now suppose two verticalperpendicular walls form an inside cornerthat you can see. Some of the sound youradiate horizontally will be headedgenerally toward the corner. It will reflectfrom both walls with high efficiency toreverse in direction and come back to you.You can stand anywhere reasonably faraway to hear a retroreflected echo of soundyou produce.

If the two walls are notperpendicular, the inside corner will notproduce retroreflection. You will generallyhear no echo of your shout or clap.

If two perpendicular walls have areasonably narrow gap between them atthe corner, you can still hear a clear echo. Itis not the corner line itself that retroreflectsthe sound, but the perpendicular walls onboth sides of the corner. Diagram (b)applies also in this case.

(a)

(b)

FIG. Q35.4

Chapter 35 317

Q35.6 The stealth fighter is designed so that adjacent panels are not joined at right angles, to prevent anyretroreflection of radar signals. This means that radar signals directed at the fighter will not bechanneled back toward the detector by reflection. Just as with sound, radar signals can be treated asdiverging rays, so that any ray that is by chance reflected back to the detector will be too weak inintensity to distinguish from background noise. This author is still waiting for the automotiveindustry to utilize this technology.

Q35.7 An echo is an example of the reflection of sound. Hearing the noise of a distant highway on a coldmorning, when you cannot hear it after the ground warms up, is an example of acoustical refraction.You can use a rubber inner tube inflated with helium as an acoustical lens to concentrate sound inthe way a lens can focus light. At your next party, see if you can experimentally find theapproximate focal point!

Q35.8 No. If the incidence angle is zero, then the ray does not change direction. Also, if the ray travels froma medium of relatively high index of refraction to one of lower index of refraction, it will bend awayfrom the normal.

Q35.9 Suppose the light moves into a medium of higher refractive index. Then its wavelength decreases.The frequency remains constant. The speed diminishes by a factor equal to the index of refraction.

Q35.10 If a laser beam enters a sugar solution with a concentration gradient (density and index of refractionincreasing with depth) then the laser beam will be progressively bent downward (toward thenormal) as it passes into regions of greater index of refraction.

Q35.11 As measured from the diagram, the incidence angle is 60°, and the refraction angle is 35°. Using

equation 35.3, sinsin

θθ

2

1

2

1=

vv

, then sinsin

3560

2°°=

vc

and the speed of light in Lucite is 2 0 108. × m s .

The frequency of the light does not change upon refraction. Knowing the wavelength in avacuum, we can use the speed of light in a vacuum to determine the frequency: c f= λ , thus

3 00 10 632 8 108 9. .× = × −f e j , so the frequency is 474.1 THz. To find the wavelength of light in Lucite,

we use the same wave speed relation, v f= λ , so 2 0 10 4 741 108 14. .× = ×e jλ , so λLucite nm= 420 .

Q35.12 Blue light would be refracted at a smaller angle from the normal, since the index of refraction forblue light—a smaller wavelength than red light—is larger.

Q35.13 The index of refraction of water is 1.33, quite different from 1.00 for air. Babies learn that therefraction of light going through the water indicates the water is there. On the other hand, the indexof refraction of liquid helium is close to that of air, so it gives little visible evidence of its presence.

Q35.14 The outgoing beam would be a rainbow, with the different colors of light traveling parallel to eachother. The white light would undergo dispersion upon refraction into the slab, with blue lightbending towards the normal more than the red light. Upon refraction out of the block, all rays oflight would exit the slab at the same angle at which they entered the slab, but offset from each other.

Q35.15 Diamond has higher index of refraction than glass and consequently a smaller critical angle for totalinternal reflection. A brilliant-cut diamond is shaped to admit light from above, reflect it totally atthe converging facets on the underside of the jewel, and let the light escape only at the top. Glasswill have less light internally reflected.


Q35.16 Light coming up from underwater is bent away from the normal. Therefore the part of the oar that issubmerged appears bent upward.

Q35.17 Highly silvered mirrors reflect about 98% of the incident light. With a 2-mirror periscope, that resultsin approximately a 4% decrease in intensity of light as the light passes through the periscope. Thismay not seem like much, but in low-light conditions, that lost light may mean the differencebetween being able to distinguish an enemy armada or an iceberg from the sky beyond. Usingprisms results in total internal reflection, meaning that 100% of the incident light is reflected throughthe periscope. That is the “total” in total internal reflection.

Q35.18 Sound travels faster in the warmer air, and thus the sound traveling through the warm air aloft willrefract much like the light refracting through the nonuniform sugar-water solution in Question35.10. Sound that would normally travel up over the tree-tops can be refracted back towards theground.

Q35.19 The light with the greater change in speed will have the larger deviation. If the glass has a higherindex than the surrounding medium, X travels slower in the glass.

Q35.20 Immediately around the dark shadow of my head, I see a halo brighter than the rest of the dewygrass. It is called the heiligenschein. Cellini believed that it was a miraculous sign of divine favorpertaining to him alone. Apparently none of the people to whom he showed it told him that theycould see halos around their own shadows but not around Cellini’s. Thoreau knew that each personhad his own halo. He did not draw any ray diagrams but assumed that it was entirely natural.Between Cellini’s time and Thoreau’s, the Enlightenment and Newton’s explanation of the rainbowhad happened. Today the effect is easy to see, whenever your shadow falls on a retroreflectingtraffic sign, license plate, or road stripe. When a bicyclist’s shadow falls on a paint stripe marking theedge of the road, her halo races along with her. It is a shame that few people are sufficiently curiousobservers of the natural world to have noticed the phenomenon.

Q35.21 Suppose the Sun is low in the sky and an observer faces away from the Sun toward a large uniformrain shower. A ray of light passing overhead strikes a drop of water. The light is refracted first at thefront surface of the drop, with the violet light deviating the most and the red light the least. At theback of the drop the light is reflected and it returns to the front surface where it again undergoesrefraction with additional dispersion as it moves from water into air. The rays leave the drop so thatthe angle between the incident white light and the most intense returning violet light is 40°, and theangle between the white light and the most intense returning red light is 42°. The observer can see aring of raindrops shining violet, a ring with angular radius 40° around her shadow. From the locus ofdirections at 42° away from the antisolar direction, the observer receives red light. The other spectralcolors make up the rainbow in between. An observer of a rainbow sees violet light at 40° angularseparation from the direction opposite the Sun, then the other spectral colors, and then red light onthe outside the rainbow, with angular radius 42°.

Q35.22 At the altitude of the plane the surface of the Earth need not block off the lower half of the rainbow.Thus, the full circle can be seen. You can see such a rainbow by climbing on a stepladder above agarden sprinkler in the middle of a sunny day. Set the sprinkler for fine mist. Do not let the slipperychildren fall from the ladder.

Q35.23 Total internal reflection occurs only when light moving originally in a medium of high index ofrefraction falls on an interface with a medium of lower index of refraction. Thus, light moving fromair (n = 1) to water (n = 1 33. ) cannot undergo total internal reflection.

Chapter 35 319

Q35.24 A mirage occurs when light changes direction as it moves between batches of air having differentindices of refraction because they have different densities at different temperatures. When the sunmakes a blacktop road hot, an apparent wet spot is bright due to refraction of light from the brightsky. The light, originally headed a little below the horizontal, always bends up as it first enters andthen leaves sequentially hotter, lower-density, lower-index layers of air closer to the road surface.

SOLUTIONS TO PROBLEMS

Section 35.1 The Nature of Light

Section 35.2 Measurements of the Speed of Light

P35.1 The Moon’s radius is 1 74 106. × m and the Earth’s radius is 6 37 106. × m . The total distance traveledby the light is:

d = × − × − × = ×2 3 84 10 1 74 10 6 37 10 7 52 108 6 6 8. . . . m m m me j .

This takes 2.51 s, so v =×

= × =7 52 10

2 995 10 299 58

8.. .

m2.51 s

m s Mm s .

P35.2 ∆x ct= ; cxt

= =×

= × =∆ 2 1 50 10 1 000

22 0 60 02 27 10 227

88

.

. ..

km m km

min s min m s Mm s

e jb ga fb g

P35.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through

one notch and returning light through the next: tc

=2

θ ω ω= = FHGIKJt

c2

so ωθ π

= =×

×=

c2

2 998 10 2 720

2 11 45 10114

8

3

.

.

e j a fe j

rad s .

The returning light would be blocked by a tooth at one-half the angular speed, giving another datapoint.

P35.4 (a) For the light beam to make it through both slots, the time for the light to travel the distance dmust equal the time for the disk to rotate through the angle θ, if c is the speed of light,

dc=θω

, so cd

=ωθ

.

(b) We are given that

d = 2 50. m , θπ

=°

°FHGIKJ = × −1 00

60 02 91 10 4.

..

rad180

rad , ωπ

= FHG

IKJ = ×5 555

23 49 104 rev s

rad1.00 rev

rad s.

cd

= =×

×= × =−

ωθ

2 50 3 49 10

2 91 103 00 10 300

4

48

. .

..

m rad s

rad m s Mm s

a fe j


Section 35.3 The Ray Approximation in Geometric Optics

Section 35.4 Reflection

Section 35.5 Refraction

*P35.5 (a) Let AB be the originally horizontal ceiling, BC its originally verticalnormal, AD the new ceiling and DE its normal. Then angleBAD = φ . By definition DE is perpendicular to AD and BC isperpendicular to AB. Then the angle between DE extended and BCis φ because angles are equal when their sides are perpendicular,right side to right side and left side to left side.

A B

C

D

E

FIG. P35.5(a)

(b) Now CBE = φ is the angle of incidence of the vertical light beam. Itsangle of reflection is also φ. The angle between the vertical incidentbeam and the reflected beam is 2φ .

A D

E

FIG. P35.5(b)

(c) tan.

.21 40

0 001 94φ = = cm

720 cmφ = °0 055 7.

P35.6 (a) From geometry, 1 25 40 0. sin . m = °d

so d = 1 94. m .

(b) 50 0. ° above the horizontal

or parallel to the incident ray.

Mirror 2

50°

40°

50°50°

40°d

i1= 40°

i2= 50°

Mirror 11.25 m

40.0°

1.25 mMirror 1

Mirror 2 Lightbeam

P

FIG. P35.6

*P35.7 (a) Method One:The incident ray makes angle α θ= °−90 1

with the first mirror. In the picture, the law of reflection impliesthat

θ θ1 1= ′ .

Then β θ θ α= °− ′ = − =90 901 1 . FIG. P35.7

In the triangle made by the mirrors and the ray passing between them,

β γγ β+ °+ = °= °−

90 18090

Further, δ γ β α= °− = =90and ∈= =δ α .

Thus the final ray makes the same angle with the first mirror as did the incident ray. Itsdirection is opposite to the incident ray.

continued on next page

Chapter 35 321

Method Two:The vector velocity of the incident light has a component vy perpendicular to the first mirror

and a component vx perpendicular to the second. The vy component is reversed upon the

first reflection, which leaves vx unchanged. The second reflection reverses vx and leaves vy

unchanged. The doubly reflected ray then has velocity opposite to the incident ray.

(b) The incident ray has velocity v v vx y zi j k+ + . Each reflection reverses one component and

leaves the other two unchanged. After all the reflections, the light has velocity

− − −v v vx y zi j k , opposite to the incident ray.

P35.8 The incident light reaches the left-hand mirror at distance

1 00 5 00 0 087 5. tan . . m ma f °=

above its bottom edge. The reflected light first reaches theright-hand mirror at height

2 0 087 5 0 175. . m mb g = .

It bounces between the mirrors with this distance betweenpoints of contact with either.

Since1 00

5 72.

. m

0.175 m=

Mirror Mirror

reflected beam1.00 m

5.00°

1.00 m

FIG. P35.8

the light reflects five times from the right-hand mirror and six times from the left .

*P35.9 Let d represent the perpendicular distance from theperson to the mirror. The distance between lamp andpeson measured parallel to the mirror can be written intwo ways: 2d d dtan tan tanθ θ φ+ = . The condition on the

distance traveled by the light is 2 2d d d

cos cos cosφ θ θ= + . We

have the two equations 3 tan tanθ φ= and2 3cos cosθ φ= . To eliminate φ we write

9 2

2

2

2sin

cossincos

θθ

φφ

= 4 92 2cos cosθ φ=

2d

d d

φ

2d tanθ θ

d tanθθd tanφ

FIG. P35.9

9 1

4 149

4 149

1 36 9 4 4

532

23 3

2 2 2 2

2 2 2 2

2 2 2 2

2

cos sin cos cos

cos sin cos cos

sin sin sin sin

sin .

φ θ θ φ

θ θ θ θ

θ θ θ θ

θ θ

= −

= −FHGIKJ

= − − = − +

= = °

e j

e j


P35.10 Using Snell’s law, sin sinθ θ21

21=

nn

θ 2 25 5= °.

λλ

21

1442= =

n nm .

FIG. P35.10

*P35.11 The law of refraction n n1 1 2 2sin sinθ θ= can be put into the more general form

cv

cv

v v

11

22

1

1

2

2

sin sin

sin sin

θ θ

θ θ

=

=

In this form it applies to all kinds of waves that move through space.

sin . sin

sin .

.

3 5343 1 493

0 266

15 4

2

2

2

°=

=

= °

m s m sθ

θ

θ

The wave keeps constant frequency in

fv v

vv

= =

= = =

1

1

2

2

22 1

1

1 493 0 589343

2 56

λ λ

λλ m s m

m s m

..

a f

P35.12 (a) fc

= =×

×= ×−λ

3 00 106 328 10

4 74 108

714.

..

m s m

Hz

(b) λλ

glassair nm

1.50 nm= = =

n632 8

422.

(c) vcnglassair m s

m s Mm s= =×

= × =3 00 10

1 502 00 10 200

88.

..

P35.13 n n1 1 2 2sin sinθ θ=

sin . sin

sin . . .

. .

θθ

θ

1

1

1

1 333 45

1 33 0 707 0 943

70 5 19 5

= °

= =

= °→ °

a fa f above the horizon

FIG. P35.13

*P35.14 We find the angle of incidence:

n n1 1 2 2

1

1

1 333 1 52 19 622 5

sin sin. sin . sin .

.

θ θθ

θ

== °

= °

The angle of reflection of the beam in water is then also 22 5. ° .

Chapter 35 323

P35.15 (a) n n1 1 2 2sin sinθ θ=

1 00 30 0 19 24

1 52

. sin . sin .

.

°= °

=

n

n

(c) fc

= =×

×= ×−λ

3 00 106 328 10

4 74 108

714.

..

m s m

Hz in air and in syrup.

(d) vcn

= =×

= × =3 00 10

1 521 98 10 198

88.

..

m s m s Mm s

(b) λ = =×

×=

vf

1 98 104 74 10

4178

14

..

m ss

nm

P35.16 (a) Flint Glass: vcn

= =×

= × =3 00 10

1 661 81 10 181

88.

..

m s m s Mm s

(b) Water: vcn

= =×

= × =3 00 10

1 3332 25 10 225

88.

..

m s m s Mm s

(c) Cubic Zirconia: vcn

= =×

= × =3 00 10

2 201 36 10 136

88.

..

m s m s Mm s

P35.17 n n1 1 2 2sin sinθ θ= : 1 333 37 0 25 02. sin . sin .°= °n

ncv2 1 90= =. : v

c= = × =

1 901 58 10 1588

.. m s Mm s

P35.18 sin sinθ θ1 2= nw

sin.

sin.

sin . . .

sin . .

tan.

.

θ θ

θ

θ

2 1

21

2

11 333

11 333

90 0 28 0 0 662

0 662 41 53 00

3 39

= = °− ° =

= = °

= =°=

−

a fa f

hd m

tan41.5 m

water

air

28°

3.0 m

h

n = 1.00

n = 1.333

θ = 62°1

θ2

FIG. P35.18


P35.19 n n1 1 2 2sin sinθ θ= : θθ

21 1 1

2=FHG

IKJ

−sinsinnn

θ 21 1 00 30

1 5019 5=

°RSTUVW = °−sin

. sin.

.

θ 2 and θ 3 are alternate interior angles formed by the ray cuttingparallel normals.

So, θ θ3 2 19 5= = °.

1 50 1 00

30 03 4

4

. sin . sin

.

θ θ

θ

=

= °

FIG. P35.19

*P35.20 For α β+ = °90

with ′ + + + = °θ α β θ1 2 180

we have ′ + = °θ θ1 2 90 .

Also, ′ =θ θ1 1

and 1 1 2sin sinθ θ= n .

Then, sin sin cosθ θ θ1 1 190= − =n nb g FIG. P35.20

sincos

tanθθ

θ1

11= =n θ 1

1= −tan n .

P35.21 At entry, n n1 1 2 2sin sinθ θ=

or 1 00 30 0 1 50 2. sin . . sin°= θ

θ 2 19 5= °. .

The distance h the light travels in the medium is given by

cos.

θ 22 00

= cm

h

or h =°=

2 002 12

..

cmcos19.5

cm .

FIG. P35.21

The angle of deviation upon entry is α θ θ= − = °− °= °1 2 30 0 19 5 10 5. . . .

The offset distance comes from sinα =dh

: d = °=2 21 10 5 0 388. sin . . cm cma f .

P35.22 The distance, h, traveled by the light is h =°=

2 002 12

..

cmcos19.5

cm..

The speed of light in the material is vch

= =×

= ×3 00 10

1 502 00 10

88.

..

m s m s

..

Therefore, thv

= =××

= × =−

−2 12 101 06 10 106

210.

. m

2.00 10 m s s ps8 .

Chapter 35 325

P35.23 Applying Snell’s law at the air-oil interface,

n nair oilsin sin .θ = °20 0

yields θ = °30 4. .

Applying Snell’s law at the oil-water interface

n nw sin sin .′ = °θ oil 20 0

yields ′ = °θ 22 3. .

FIG. P35.23

*P35.24 For sheets 1 and 2 as described,

n nn n

1 2

1 2

26 5 31 70 849

sin . sin ..

°= °=

For the trial with sheets 3 and 2,

n nn n

3 2

3 2

26 5 36 70 747

sin . sin ..

°= °=

Now

0 747 0 8491 14

3 1

3 1

. ..

n nn n

==

For the third trial,

n n n1 3 3 1 3

3

26 5 1 14

23 1

sin . sin . sin

.

°= =

= °

θ θ

θ

P35.25 Consider glass with an index of refraction of 1.5, which is 3 mm thick. The speed of light in the glassis

3 101 5

2 108

8×= ×

m s m s

..

The extra travel time is3 102 10

3 103 10

103

8

3

811×

×−

××

− −− m

m s m

m s s~ .

For light of wavelength 600 nm in vacuum and wavelength 600

400 nm

1.5 nm= in glass,

the extra optical path, in wavelengths, is3 10 3 10

6 1010

3

7

3

73×

×−

××

−

−

−

− m

4 10 m m m

wavelengths~ .


P35.26 Refraction proceeds according to 1 00 1 661 2. sin . sina f a fθ θ= . (1)

(a) For the normal component of velocity to be constant, v v1 1 2 2cos cosθ θ=

or cca fcos.

cosθ θ1 21 66= FHGIKJ . (2)

We multiply Equations (1) and (2), obtaining: sin cos sin cosθ θ θ θ1 1 2 2=

or sin sin2 21 2θ θ= .

The solution θ θ1 2 0= = does not satisfy Equation (2) and must be rejected. The physicalsolution is 2 180 21 2θ θ= °− or θ θ2 190 0= °−. . Then Equation (1) becomes:

sin . cosθ θ1 11 66= , or tan .θ 1 1 66=

which yields θ 1 58 9= °. .

(b) Light entering the glass slows down and makes a smaller angle with the normal. Both effectsreduce the velocity component parallel to the surface of the glass, so that component cannotremain constant, or will remain constant only in the trivial case θ θ1 2 0= = .

P35.27 See the sketch showing the path of thelight ray. α and γ are angles of incidence atmirrors 1 and 2.

For triangle abca,

2 2 180α γ β+ + = °

or β α γ= °− +180 2b g . (1)

Now for triangle bcdb,

90 0 90 0 180. .°− + °− + = °α γ θa f b gor θ α γ= + . (2)

FIG. P35.27

Substituting Equation (2) into Equation (1) gives β θ= °−180 2 .

Note: From Equation (2), γ θ α= − . Thus, the ray will follow a path like that shown only if α < 0 .For α > 0 , γ is negative and multiple reflections from each mirror will occur before the incident andreflected rays intersect.

Chapter 35 327

Section 35.6 Huygen’s Principle

*P35.28 (a) For the diagrams of contour lines and wave fronts and rays, see Figures (a) and (b) below.As the waves move to shallower water, the wave fronts bend to become more nearly parallelto the contour lines.

(b) For the diagrams of contour lines and wave fronts and rays, see Figures (c) and (d) below.We suppose that the headlands are steep underwater, as they are above water. The rays areeverywhere perpendicular to the wave fronts of the incoming refracting waves. As shown,the rays bend toward the headlands and deliver more energy per length at the headlands.

(a) Contour lines (b) Wave fronts (c) Contour lines (d) Wave frontsand rays and rays

FIG. P35.28

Section 35.7 Dispersion and Prisms

P35.29 From Fig 35.21 nv = 1 470. at 400 nm and nr = 1 458. at 700 nm.

Then 1 00 1 470. sin . sinθ θ= v and 1 00 1 458. sin . sinθ θ= r

δ δ θ θθ θ

δ

r v r v− = − = FHGIKJ −

FHGIKJ

=°F

HGIKJ −

°FHG

IKJ = °

− −

− −

sinsin.

sinsin.

sinsin .

.sin

sin ..

.

1 1

1 1

1 458 1 47030 0

1 45830 0

1 4700 171∆

P35.30 n 700 1 458 nma f = .

(a) 1 00 75 0 1 458 2. sin . . sina f °= θ ; θ 2 41 5= °.

(b) Let θ β3 90 0+ = °. , θ α2 90 0+ = °. then α β+ + °= °60 0 180. .

So 60 0 0 60 0 41 5 18 52 3 3. . . .°− − = ⇒ °− °= = °θ θ θ .

(c) 1 458 18 5 1 00 4. sin . . sin°= θ θ 4 27 6= °.

(d) γ θ θ β θ= − + − °−1 2 490 0b g b g.

γ = °− °+ °− ° − °− ° = °75 0 41 5 90 0 18 5 90 0 27 6 42 6. . . . . . .a f a f

60.0°γα

θ1 θ2 θ3

θ4β

FIG. P35.30


P35.31 Taking Φ to be the apex angle and δ min to be the angle of minimum deviation, from Equation 35.9,the index of refraction of the prism material is

n =+sin

sinminΦ

Φ

δb gb g

2

2

Solving for δ min , δ min sin sin sin . sin . . .= FHG

IKJ − = ° − °= °− −2

22 2 20 25 0 50 0 86 81 1n

ΦΦ a f a f .

P35.32 Note for use in every part: Φ+ °− + °− = °90 0 90 0 1802 3. .θ θb g b gso θ θ3 2= −Φ .

At the first surface the deviation is α θ θ= −1 2 .

At exit, the deviation is β θ θ= −4 3 .

The total deviation is therefore δ α β θ θ θ θ θ θ= + = + − − = + −1 4 2 3 1 4 Φ . FIG. P35.32

(a) At entry: n n1 1 2 2sin sinθ θ= or θ 21 48 6

1 5030 0=

°FHG

IKJ = °−sin

sin ..

. .

Thus, θ 3 60 0 30 0 30 0= °− °= °. . . .

At exit: 1 50 30 0 1 00 4. sin . . sin°= θ or θ 41 1 50 30 0 48 6= ° = °−sin . sin . .a f

so the path through the prism is symmetric when θ 1 48 6= °. .

(b) δ = °+ °− °= °48 6 48 6 60 0 37 2. . . .

(c) At entry: sinsin .

..θ θ2 2

45 61 50

28 4=°⇒ = ° θ 3 60 0 28 4 31 6= °− °= °. . . .

At exit: sin . sin . .θ θ4 41 50 31 6 51 7= ° ⇒ = °a f δ = °+ °− °= °45 6 51 7 60 0 37 3. . . . .

(d) At entry: sinsin .

..θ θ2 2

51 61 50

31 5=°⇒ = ° θ 3 60 0 31 5 28 5= °− °= °. . . .

At exit: sin . sin . .θ θ4 41 50 28 5 45 7= ° ⇒ = °a f δ = °+ °− °= °51 6 45 7 60 0 37 3. . . . .

P35.33 At the first refraction, 1 00 1 2. sin sinθ θ= n .

The critical angle at the second surface is given by n sin .θ 3 1 00= :

or θ 31 1 00

1 5041 8= F

HGIKJ = °−sin

.

.. .

But, θ θ2 360 0= °−. .FIG. P35.33

Thus, to avoid total internal reflection at the second surface (i.e., have θ 3 41 8< °. )

it is necessary that θ 2 18 2> °. .

Since sin sinθ θ1 2= n , this becomes sin . sin . .θ 1 1 50 18 2 0 468> °=

or θ 1 27 9> °. .

Chapter 35 329

P35.34 At the first refraction, 1 00 1 2. sin sinθ θ= n .The critical angle at the second surface is given by

n sin .θ 3 1 00= , or θ 31 1 00

= FHGIKJ

−sin.n

.

But 90 0 90 0 1802 3. .°− + °− + = °θ θb g b g Φ

which gives θ θ2 3= −Φ .

Φ

θ2

θ1

θ3

FIG. P35.34

Thus, to have θ 31 1 00

< FHGIKJ

−sin.n

and avoid total internal reflection at the second surface,

it is necessary that θ 21 1 00

> − FHGIKJ

−Φ sin.n

.

Since sin sinθ θ1 2= n , this requirement becomes sin sin sin.

θ 11 1 00

> − FHGIKJ

LNM

OQP

−nn

Φ

or θ 11 1 1 00

> − FHGIKJ

LNM

OQP

FHG

IKJ

− −sin sin sin.

nn

Φ .

Through the application of trigonometric identities, θ 11 2 1> − −FH IK−sin sin cosn Φ Φ .

P35.35 For the incoming ray, sinsin

θθ

21=

n.

Using the figure to the right, θ 21 50 0

1 6627 48b gviolet

=°F

HGIKJ = °−sin

sin ..

.

θ 21 50 0

1 6228 22b gred

=°F

HGIKJ = °−sin

sin ..

. .

For the outgoing ray, θ θ3 260 0= °−.FIG. P35.35

and sin sinθ θ4 3= n : θ 41 1 66 32 52 63 17b gviolet

= ° = °−sin . sin . .

θ 41 1 62 31 78 58 56b gred

= ° = °−sin . sin . . .

The angular dispersion is the difference ∆θ θ θ4 4 4 63 17 58 56 4 61= − = °− °= °b g b gviolet red. . . .

Section 35.8 Total Internal Reflection

P35.36 n sinθ = 1 . From Table 35.1,

(a) θ = FHGIKJ = °−sin

..1 1

2 41924 4

(b) θ = FHGIKJ = °−sin

..1 1

1 6637 0

(c) θ = FHGIKJ = °−sin

..1 1

1 30949 8


P35.37 sinθ cnn

= 2

1: θ c

nn

=FHGIKJ

−sin 1 2

1

(a) Diamond: θ c =FHGIKJ = °−sin

.

..1 1 333

2 41933 4

(b) Flint glass: θ c =FHGIKJ = °−sin

..

.1 1 3331 66

53 4

(c) Ice: Since n n2 1> , there is no critical angle .

P35.38 sin..

.θ cn

n= = =air

pipe

1 001 36

0 735 θ c = °47 3.

Geometry shows that the angle of refraction at the end isφ θ= °− = °− °= °90 0 90 0 47 3 42 7. . . .c .

Then, Snell’s law at the end, 1 00 1 36 42 7. sin . sin .θ = °

gives θ = °67 2. .

The 2-µm diameter is unnecessary information.

FIG. P35.38

P35.39 sinθ cnn

= 2

1

n n2 1 88 8 1 000 3 0 999 8 1 000 08= °= =sin . . . .b gb gFIG. P35.39

*P35.40 (a) A ray along the inner edge will escape if any ray escapes. Its angle of

incidence is described by sinθ =−R dR

and by n sin sinθ > °1 90 . Then

n R dR−

>a f

1 nR nd R− > nR R nd− > Rnd

n>

− 1.

(b) As d → 0 , Rmin → 0 . This is reasonable.As n increases, Rmin decreases. This is reasonable.As n decreases toward 1, Rmin increases. This is reasonable.

FIG. P35.40

(c) Rmin

.

.=

×= ×

−−

1 40 100 10

0 40350 10

66

m m

e j

P35.41 From Snell’s law, n n1 1 2 2sin sinθ θ= .At the extreme angle of viewing, θ 2 90 0= °.

1 59 1 00 90 01. sin . sin .a fb g a fθ = ° .

So θ 1 39 0= °. .

Therefore, the depth of the air bubble is

rd

rd p

tan tanθ θ1 1< <

or 1 08 1 17. . cm cm< <d . FIG. P35.41

Chapter 35 331

P35.42 (a)sinsin

θθ

2

1

2

1=

vv

and θ 2 90 0= °. at the critical angle

sin .sin

90 0 1 850343

°=

θ c

m s m s

so θ c = = °−sin . .1 0 185 10 7a f .

(b) Sound can be totally reflected if it is traveling in the medium where it travels slower: air .

(c) Sound in air falling on the wall from most directions is 100% reflected , so the wall is a

good mirror.

P35.43 For plastic with index of refraction n ≥ 1 42. surrounded by air, the critical angle for total internal

reflection is given by

θ c n= FHGIKJ ≤

FHGIKJ = °− −sin sin

..1 11 1

1 4244 8 .

In the gasoline gauge, skylight from above travels down the plastic. The rays close to the vertical aretotally reflected from the sides of the slab and from both facets at the lower end of the plastic, whereit is not immersed in gasoline. This light returns up inside the plastic and makes it look bright.Where the plastic is immersed in gasoline, with index of refraction about 1.50, total internalreflection should not happen. The light passes out of the lower end of the plastic with little reflected,making this part of the gauge look dark. To frustrate total internal reflection in the gasoline, theindex of refraction of the plastic should be n < 2 12. .

since θ c =FHGIKJ = °−sin

.

..1 1 50

2 1245 0 .

Section 35.9 Fermat‘s Principle

P35.44 Assume the lifeguard’s initial path makes angle θ 1 with the north-south normal to the shoreline, and angle θ 2 with this normal inthe water. By Fermat’s principle, his path should follow the law ofrefraction:sinsin

.

..

θθ

1

2

1

2

7 001 40

5 00= = =vv

m s m s

or θθ

21 1

5= FHGIKJ

−sinsin

.FIG. P35.44

The lifeguard on land travels eastward a distance x = 16 0 1. tan ma f θ . Then in the water, he travels

26 0 20 0 2. . tan m m− =x a f θ further east. Thus, 26 0 16 0 20 01 2. . tan . tan m m m= +a f a fθ θ

or 26 0 16 0 20 051

1 1. . tan . tan sinsin

m m m= + FHGIKJ

LNM

OQP

−a f a fθθ

.

We home in on the solution as follows:

θ 1 (deg) 50.0 60.0 54.0 54.8 54.81

right-hand side 22.2 m 31.2 m 25.3 m 25.99 m 26.003 m

The lifeguard should start running at 54 8. ° east of north .


Additional Problems

*P35.45 Scattered light leaves the center of the photograph (a) in allhorizontal directions between θ 1 0= ° and 90° from thenormal. When it immediately enters the water (b), it isgathered into a fan between 0° and θ 2 max given by

n n1 1 2 2

2

2

1 00 90 1 33348 6

sin sin. sin . sin

.

θ θθ

θ

==

= ° max

max

The light leaves the cylinder without deviation, so theviewer only receives light from the center of thephotograph when he has turned by an angle less than48.6°. When the paperweight is turned farther, light at theback surface undergoes total internal reflection (c). Theviewer sees things outside the globe on the far side.

θ 1 max

(a)

θ 2 max

(b)

(c)

FIG. P35.45

P35.46 Let n xa f be the index of refraction at distance x below the top of the atmosphere and n x h n= =a f beits value at the planet surface.

Then, n xn

hxa f = +

−FHG

IKJ1 000

1 000.

..

(a) The total time interval required to traverse the atmosphere is

∆tdxv

n xc

dxh h

= =z z0 0

a f: ∆t

cn

hx dx

h

= +−FHG

IKJ

LNM

OQPz1 1 000

1 000

0

..

∆thc

nch

h hc

n= +

− FHGIKJ =

+FHG

IKJ

1 0002

1 0002

2. .a f.

(b) The travel time in the absence of an atmosphere would be hc

.

Thus, the time in the presence of an atmosphere is n +FHG

IKJ

1 0002.

times larger .

P35.47 Let the air and glass be medium 1 and 2, respectively. By Snell’s law, n n2 2 1 1sin sinθ θ=

or 1 56 2 1. sin sinθ θ= .

But the conditions of the problem are such that θ θ1 22= . 1 56 22 2. sin sinθ θ= .

We now use the double-angle trig identity suggested. 1 56 22 2 2. sin sin cosθ θ θ=

or cos.

.θ 21 56

20 780= = .

Thus, θ 2 38 7= °. and θ θ1 22 77 5= = °. .

Chapter 35 333

P35.48 (a) ′ = = °θ θ1 1 30 0. n n1 1 2 2sin sinθ θ=

1 00 30 0 1 55

18 82

2

. sin . . sin

.

°=

= °

θ

θ

(b) ′ = = °θ θ1 1 30 0. θθ

21 1 1

2

1 1 55 30 01

50 8

=FHG

IKJ

=°F

HGIKJ = °

−

−

sinsin

sin. sin .

.

nn

FIG. P35.48

(c), (d) The other entries are computed similarly, and are shown in the table below.

(c) air into glass, angles in degrees (d) glass into air, angles in degrees

incidence reflection refraction incidence reflection refraction0 0 0 0 0 0

10.0 10.0 6.43 10.0 10.0 15.620.0 20.0 12.7 20.0 20.0 32.030.0 30.0 18.8 30.0 30.0 50.840.0 40.0 24.5 40.0 40.0 85.150.0 50.0 29.6 50.0 50.0 none*60.0 60.0 34.0 60.0 60.0 none*70.0 70.0 37.3 70.0 70.0 none*80.0 80.0 39.4 80.0 80.0 none*90.0 90.0 40.2 90.0 90.0 none*

*total internal reflection

P35.49 For water, sinθ c = =1

4 334

.

Thus θ c = = °−sin . .1 0 750 48 6a fand d c= 2 1 00. tan ma f θ

d = °=2 00 48 6 2 27. tan . . m ma f . FIG. P35.49

P35.50 Call θ 1 the angle of incidence and of reflection on theleft face and θ 2 those angles on the right face. Let αrepresent the complement of θ 1 and β be thecomplement of θ 2 . Now α γ= and β δ= becausethey are pairs of alternate interior angles. We have

A = + = +γ δ α β

and B A A A= + + = + + =α β α β 2 .

FIG. P35.50


P35.51 (a) We see the Sun moving from east to west across the sky. Its angular speed is

ωθ π

= = = × −∆∆t

27 27 10 5 rad

86 400 s rad s. .

The direction of sunlight crossing the cell from the window changes at this rate, moving onthe opposite wall at speed

v r= = × = × =− −ω 2 37 7 27 10 1 72 10 0 1725 4. . . . m rad s m s mm sa fe j .

(b) The mirror folds into the cell the motion that would occur in a room twice as wide:

v r= = =ω 2 0 174 0 345. . mm s mm sb g .

(c), (d) As the Sun moves southward and upward at 50.0°, we may regard the corner of the windowas fixed, and both patches of light move northward and downward at 50.0° .

*P35.52 (a) 45 0. ° as shown in the first figure to the right.

(b) Yes

If grazing angle is halved, the number of reflections fromthe side faces is doubled. FIG. P35.52

P35.53 Horizontal light rays from the setting Sun passabove the hiker. The light rays are twice refractedand once reflected, as in Figure (b). The mostintense light reaching the hiker, that whichrepresents the visible rainbow, is located betweenangles of 40° and 42° from the hiker’s shadow.

The hiker sees a greater percentage of the violetinner edge, so we consider the red outer edge. Theradius R of the circle of droplets is

R = °=8 00 42 0 5 35. sin . . km kma f .

Then the angle φ, between the vertical and theradius where the bow touches the ground, is givenby

cos. .

.φ = = =2 00 2 00

0 374 km km

5.35 kmR

or φ = °68 1. .

The angle filled by the visible bow is360 2 68 1 224°− × ° = °.a fso the visible bow is

224360

62 2%°°= . of a circle .

Figure (a)

Figure (b)

FIG. P35.53

Chapter 35 335

P35.54 Light passing the top of the pole makes an angle of incidenceφ θ1 90 0= °−. . It falls on the water surface at distance from the pole

sL d

1 =−

tanθand has an angle of refraction φ 2 from 1 00 1 2. sin sinφ φ= n .Then s d2 2= tanφand the whole shadow length is

s sL d

dn1 2

1 1+ =−

+ FHGIKJ

FHG

IKJ

−

tantan sin

sinθ

φ

s sL d

dn1 2

1

12 0040 0

2 0040 0

1 333 79

+ =−

+ FHGIKJ

FHG

IKJ

=°+

°FHG

IKJ

FHG

IKJ =

−

−

tantan sin

cos

.tan .

. tan sincos .

..

θθ

m m ma f FIG. P35.54

P35.55 As the beam enters the slab,

1 00 50 0 1 48 2. sin . . sin°= θ

giving θ 2 31 2= °. .FIG. P35.55

The beam then strikes the top of the slab at x11 55

31 2=

°.

tan . mm

from the left end. Thereafter, the beam

strikes a face each time it has traveled a distance of 2 1x along the length of the slab. Since the slab is420 mm long, the beam has an additional 420 1 mm − x to travel after the first reflection. The numberof additional reflections is

4202

1 55 31 23 10 31 2

81 51

1

mm 420 mm mm mm

−=

− °°

=x

x. tan .

. tan .. or 81 reflections

since the answer must be an integer. The total number of reflections made in the slab is then 82 .

P35.56 (a)′=

−+

LNM

OQP

=−+

LNM

OQP =

SS

n nn n

1

1

2 1

2 1

2 21 52 1 001 52 1 00

0 042 6. .. .

.

(b) If medium 1 is glass and medium 2 is air,′=

−+

LNM

OQP

=−+

LNM

OQP =

SS

n nn n

1

1

2 1

2 1

2 21 00 1 521 00 1 52

0 042 6. .. .

. .

There is no difference .

P35.57 (a) With n1 1=and n n2 =

the reflected fractional intensity is′=

−+FHGIKJ

SS

nn

1

1

211

.

The remaining intensity must be transmitted:

SS

nn

n n

n

n n n n

n

n

n2

1

2 2 2

2

2 2

2 2111

1 1

1

2 1 2 1

1

4

1= −

−+FHGIKJ =

+ − −

+=

+ + − + −

+=

+

a f a fa f a f a f .

continued on next page


(b) At entry,SS

nn

2

1

2

2111

4 2 419

2 419 10 828= −

−+FHGIKJ =

+=

.

..

a fa f .

At exit,SS

3

20 828= . .

Overall,SS

SS

SS

3

1

3

2

2

1

20 828 0 685=FHGIKJFHGIKJ = =. .a f

or 68 5%. .

P35.58 Define Tn

n=

+

4

1 2a f as the transmission coefficient for one

encounter with an interface. For diamond and air, it is 0.828, asin Problem 57.

As shown in the figure, the total amount transmitted is

T T T T T T T

T T n

2 2 2 2 4 2 6

2 2

1 1 1

1

+ − + − + −

+ + − +

a f a f a fa f… …

We have 1 1 0 828 0 172− = − =T . . so the total transmission is

0 828 1 0 172 0 172 0 1722 2 4 6. . . .a f a f a f a f+ + + +… .

To sum this series, define F = + + + +1 0 172 0 172 0 1722 4 6. . .a f a f a f … .

Note that 0 172 0 172 0 172 0 1722 2 4 6. . . .a f a f a f a fF = + + +… , and

1 0 172 1 0 172 0 172 0 1722 2 4 6+ = + + + + =. . . .a f a f a f a fF F… .

FIG. P35.58

Then, 1 0 172 2= −F F.a f or F =−

1

1 0 172 2.a f .

The overall transmission is then 0 828

1 0 1720 706

2

2

.

..

a fa f−

= or 70 6%. .

P35.59 Define n1 to be the index of refraction of the surrounding medium and n2 tobe that for the prism material. We can use the critical angle of 42.0° to find the

ratio nn

2

1:

n n2 142 0 90 0sin . sin .°= ° .

So,nn

2

1

142 0

1 49=°=

sin .. .

Call the angle of refraction θ 2 at the surface 1. The ray inside the prism formsa triangle with surfaces 1 and 2, so the sum of the interior angles of thistriangle must be 180°.

FIG. P35.59

Thus, 90 0 60 0 90 0 42 0 1802. . . .°− + °+ °− ° = °θb g a f .

Therefore, θ 2 18 0= °. .Applying Snell’s law at surface 1, n n1 1 2 18 0sin sin .θ = °

sin sin . sin .θ θ12

12 1 49 18 0=

FHGIKJ = °

nn

θ 1 27 5= °. .

Chapter 35 337

*P35.60 (a) As the mirror turns through angle θ, the angle of incidenceincreases by θ and so does the angle of reflection. The incident rayis stationary, so the reflected ray turns through angle 2θ . Theangular speed of the reflected ray is 2ωm . The speed of the dot of

light on the circular wall is 2ωmR .

(b) The two angles marked θ in the figure to the right are equalbecause their sides are perpendicular, right side to right side andleft side to left side.

We have cosθ =+

=d

x d

dsdx2 2

anddsdt

x dm= +2 2 2ω .

Sodxdt

dsdt

x dd

x ddm=

+=

+2 2 2 2

2ω .

FIG. P35.60

P35.61 (a) For polystyrene surrounded by air, internal reflection requires

θ 31 1 00

1 4942 2= F

HGIKJ = °−sin

.

.. .

Then from geometry, θ θ2 390 0 47 8= °− = °. . .

From Snell’s law, sin . sin . .θ 1 1 49 47 8 1 10= °= .

This has no solution.

Therefore, total internal reflection always happens .

(b) For polystyrene surrounded by water, θ 31 1 33

1 4963 2= F

HGIKJ = °−sin

.

..

and θ 2 26 8= °. .

From Snell’s law, θ 1 30 3= °. .

FIG. P35.61

(c) No internal refraction is possible

since the beam is initially traveling in a medium of lower index of refraction.


*P35.62 The picture illustrates optical sunrise. At the center of the earth,

cos.

.. .

φ

φθ

=×

× += °= − °= °

6 37 108 614

2 9890 2 98 87 0

6

2

m6.37 10 m6

At the top of the atmosphere

n n1 1 2 2

1

1

1 1 000 293 87 087 4

sin sinsin . sin .

.

θ θθ

θ

== °

= °

δ1

2

θ1

θ2

φ

FIG. P35.62

Deviation upon entry is

δ θ θδ= −

= °− °= °1 2

87 364 87 022 0 342. . .

Sunrise of the optical day is before geometric sunrise by 0 34286 400

82 2. .°°

FHG

IKJ =

s360

s. Optical sunset

occurs later too, so the optical day is longer by 164 s .

P35.63 tan.

θ 14 00

= cm

h

and tan.

θ 22 00

= cm

h

tan . tan . tan21 2

2 222 00 4 00θ θ θ= =b g

sinsin

.sin

sin

21

21

22

221

4 001

θθ

θθ−

=−

FHG

IKJ . (1)

Snell’s law in this case is: n n1 1 2 2sin sinθ θ=

sin . sinθ θ1 21 333= .

Squaring both sides, sin . sin21

221 777θ θ= . (2)

Substituting (2) into (1),1 777

1 1 7774 00

1

22

22

22

22

. sin. sin

.sin

sinθθ

θθ−

=−

FHG

IKJ .

Defining x = sin2 θ ,0 444

1 1 7771

1..−

=−x x .

FIG. P35.63

Solving for x, 0 444 0 444 1 1 777. . .− = −x x and x = 0 417. .

From x we can solve for θ 2 : θ 21 0 417 40 2= = °−sin . . .

Thus, the height is h = =°=

2 00 2 0040 2

2 362

.tan

.tan .

. cm cm

cmθ

.

Chapter 35 339

P35.64 δ θ θ= − = °1 2 10 0. and n n1 1 2 2sin sinθ θ=

with n1 1= , n243

= .

Thus, θ θ θ11

2 21

2 1 10 0= = − °− −sin sin sin sin .n nb g b g .

(You can use a calculator to home in on an approximate solution to this equation, testing differentvalues of θ 1 until you find that θ 1 36 5= °. . Alternatively, you can solve for θ 1 exactly, as shown

below.)

We are given that sin sin .θ θ1 143

10 0= − °b g .

This is the sine of a difference, so34

10 0 10 01 1 1sin sin cos . cos sin .θ θ θ= °− ° .

Rearranging, sin . cos cos . sin10 0 10 0341 1° = °−F

HGIKJθ θ

sin .cos . .

tan10 0

10 0 0 750 1°

°−= θ and θ 1

1 0 740 36 5= = °−tan . .a f .

P35.65 To derive the law of reflection, locate point O so that the time of travelfrom point A to point B will be minimum.

The total light path is L a b= +sec secθ θ1 2 .

The time of travel is tv

a b= FHGIKJ +

11 2sec secθ θb g .

If point O is displaced by dx, then

FIG. P35.65

dtv

a d b d= FHGIKJ + =

101 1 1 2 2 2sec tan sec tanθ θ θ θ θ θb g (1)

(since for minimum time dt = 0 ).

Also, c d a b+ = + =tan tanθ θ1 2 constant

so, a d b dsec sec21 1

22 2 0θ θ θ θ+ = . (2)

Divide equations (1) and (2) to find θ θ1 2= .


P35.66 Observe in the sketch that the angle of incidence at point P is γ, andusing triangle OPQ:

sinγ =LR

.

Also, cos sinγ γ= − =−

1 22 2R LR

.

Applying Snell’s law at point P, 1 00. sin sinγ φ= n .

Thus, sinsin

φγ

= =n

LnR

and cos sinφ φ= − =−

1 22 2 2n R L

nR.

FIG. P35.66

From triangle OPS, φ α γ+ + ° + °− = °90 0 90 0 180. .a f b g or the angle of incidence at point S is α γ φ= − .

Then, applying Snell’s law at point S

gives 1 00. sin sin sinθ α γ φ= = −n n b g

or sin sin cos cos sinθ γ φ γ φ= − = FHGIKJ

−−

− FHGIKJ

LNMM

OQPPn n

LR

n R LnR

R LR

LnR

2 2 2 2 2

sinθ = − − −FH IKL

Rn R L R L2

2 2 2 2 2

and θ = − − −FH IKLNM

OQP

−sin 12

2 2 2 2 2LR

n R L R L .

P35.67 As shown in the sketch, the angle of incidence at point A is:

θ = FHGIKJ =

FHG

IKJ = °− −sin sin

..1 12 1 00

30 0dR

m2.00 m

.

If the emerging ray is to be parallel to the incident ray, the pathmust be symmetric about the centerline CB of the cylinder. Inthe isosceles triangle ABC,

γ α= and β θ= °−180 .

Therefore, α β γ+ + = °180

becomes 2 180 180α θ+ °− = °

or αθ

= = °2

15 0. .

Then, applying Snell’s law at point A,

n sin . sinα θ= 1 00

or n = =°°=

sinsin

sin .sin .

.θα

30 015 0

1 93 .

FIG. P35.67

Chapter 35 341

*P35.68 (a) The apparent radius of the glowing sphereis R3 as shown. For it

sin

sin

sin sin

θ

θ

θ θ

11

2

23

2

1 2

1

2

3

23 1

1

=

=

=

= =

RRRR

n

nRR

RR

R nR

R1

R2

R3θ1

θ2

FIG. P35.68(a)

(b) If nR R1 2> , then sinθ 2 cannot be equal tonRR

1

2. The ray considered in part (a)

undergoes total internal reflection. In thiscase a ray escaping the atmosphere asshown here is responsible for the apparentradius of the glowing sphere and

R R3 2= .

R3

FIG. P35.68(b)

P35.69 (a) At the boundary of the air and glass, the critical angle is given by

sinθ c n=

1.

Consider the critical ray PBB′ : tanθ cd

t=

4 or

sincos

θθ

c

c

dt

=4

.

Squaring the last equation gives:sincos

sinsin

2

2

2

2

2

1 4θθ

θθ

c

c

c

c

dt

=−

= FHGIKJ .

tn

P

B

B'

θc

air

d

d /4

FIG. P35.69

Since sinθ c n=

1, this becomes

11 42

2

ndt−

= FHGIKJ or n

td

= + FHGIKJ1

4 2

.

(b) Solving for d, dt

n=

−

4

12.

Thus, if n = 1 52. and t = 0 600. cm , d =−

=4 0 600

1 52 12 10

2

.

..

cm cm

a fa f

.

(c) Since violet light has a larger index of refraction, it will lead to a smaller critical angle andthe inner edge of the white halo will be tinged with violet light.


P35.70 From the sketch, observe that the angle ofincidence at point A is the same as theprism angle θ at point O. Given thatθ = °60 0. , application of Snell’s law atpoint A gives

1 50 1 00 60 0. sin . sin .β = ° or β = °35 3. .

From triangle AOB, we calculate the angleof incidence (and reflection) at point B.

FIG. P35.70

θ β γ= °− + °− = °90 0 90 0 180. .b g b g so γ θ β= − = °− °= °60 0 35 3 24 7. . . .

Now, using triangle BCQ: 90 0 90 0 90 0 180. . .°− + °− + °− = °γ δ θb g b g a f .

Thus the angle of incidence at point C is δ θ γ= °− − = °− °= °90 0 30 0 24 7 5 30. . . .a f .

Finally, Snell’s law applied at point C gives 1 00 1 50 5 30. sin . sin .φ = °

or φ = ° = °−sin . sin . .1 1 50 5 30 7 96a f .

P35.71 (a) Given that θ 1 45 0= °. and θ 2 76 0= °. .

Snell’s law at the first surface gives

n sin . sin .α = °1 00 45 0 (1)

Observe that the angle of incidence at the secondsurface is

β α= °−90 0. .

Thus, Snell’s law at the second surface yields

n nsin sin . . sin .β α= °− = °90 0 1 00 76 0a for n cos sin .α = °76 0 . (2)

FIG. P35.71

Dividing Equation (1) by Equation (2), tansin .sin .

.α =°°=

45 076 0

0 729

or α = °36 1. .

Then, from Equation (1), n =°=

°°=

sin .sin

sin .sin .

.45 0 45 0

36 11 20

α.

(b) From the sketch, observe that the distance the light travels in the plastic is dL

=sinα

. Also,

the speed of light in the plastic is vcn

= , so the time required to travel through the plastic is

∆tdv

nLc

= = =× °

= × =−

sin. .

. sin .. .

α1 20 0 500

3 00 10 36 13 40 10 3 40

89 m

m s s ns

a fe j

.

Chapter 35 343

P35.72sin sin

sinsin

θ θθθ1 2

1

2

0.1740.3420.5000.6430.7660.8660.9400.985

0.1310.2610.3790.4800.5760.6470.7110.740

1.330 41.312 91.317 71.338 51.328 91.339 01.322 01.331 5

The straightness of the graph line demonstrates Snell’sproportionality.

The slope of the line is n = ±1 327 6 0 01. .

and n = ±1 328 0 8%. . .

FIG. P35.72

ANSWERS TO EVEN PROBLEMS

P35.2 227 Mm s P35.30 (a) 41 5. ° ; (b) 18 5. °; (c) 27 6. ° ; (d) 42 6. °

P35.32 (a) see the solution; (b) 37 2. ° ; (c) 37 3. ° ;P35.4 (a) see the solution; (b) 300 Mm s(d) 37 3. °

P35.6 (a) 1 94. m; (b) 50 0. ° above the horizontal :antiparallel to the incident ray P35.34 sin sin cos− − −FH IK1 2 1n Φ Φ

P35.8 five times by the right-hand mirror andsix times by the left-hand mirror

P35.36 (a) 24 4. ° ; (b) 37 0. ° ; (c) 49 8. °

P35.38 67 2.P35.10 25 5. ° ; 442 nm

P35.40 (a) nd

n − 1; (b) yes; (c) 350 mµP35.12 (a) 474 THz ; (b) 422 nm; (c) 200 Mm s

P35.14 22 5. ° P35.42 (a) 10 7. °; (b) air; (c) Sound falling on thewall from most directions is 100%reflected.P35.16 (a) 181 Mm s; (b) 225 Mm s;

(c) 136 Mm sP35.44 54 8. ° east of north

P35.18 3 39. mP35.46 (a)

hc

n +FHGIKJ

12

; (b) larger by n + 1

2 times

P35.20 θ 11= −tan n

P35.48 see the solutionP35.22 106 ps

P35.50 see the solutionP35.24 23.1°

P35.52 (a) 45 0. ° ; (b) yes; see the solutionP35.26 (a) 58 9. ° ; (b) Only if θ θ1 2 0= =

P35.54 3 79. mP35.28 see the solution


P35.56 (a) 0 042 6. ; (b) no differenceP35.66 θ = − − −FH IK

LNM

OQP

−sin 12

2 2 2 2 2LR

n R L R L

P35.58 0 706.P35.68 (a) nR1 ; (b) R2

P35.60 (a) 2ωmR ; (b) 22 2

ωmx d

d+

P35.70 7 96. °

P35.62 164 s P35.72 see the solution; n = ±1 328 0 8%. .

P35.64 36 5. °

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