Chapter 24

Wave Optics

Quick Quizzes

1.

(c). The fringes on the screen are equally spaced only at small angles where tan θ ≈ sin θ is

a valid approximation.

2.

(b). The space between successive bright fringes is proportional to the wavelength of the

light. Since the wavelength in water is less than that in air, the bright fringes are closer

together in the second experiment.

3.

(b). The outer edges of the central maximum occur where sin θ = ± λ a . Thus, as a, the

width of the slit, becomes smaller, the width of the central maximum will increase.

4.

The compact disc. The tracks of information on a compact disc are much closer together

than on a phonograph record. As a result, the diffraction maxima from the compact disc

will be farther apart than those from the record.

303

304

CHAPTER 24

Answers to Even Numbered Conceptual Questions

2.

The wavelength of light is extremely small in comparison to the dimensions of your hand,

so the diffraction of light around obstacles the size of your hand is totally negligible.

However, sound waves have wavelengths that are comparable to the dimensions of the

hand or even larger. Therefore, significant diffraction of sound waves occurs around hand

sized obstacles.

4.

The wavelength of light traveling in water would decrease, since the wavelength of light

in a medium is given by λn = λ n , where λ is the wavelength in vacuum and n is the

index of refraction of the medium. Since the positions of the bright and dark fringes are

proportional to the wavelength, the fringe separations would decrease.

6.

Every color produces its own interference pattern, and we see them superimposed. The

central maximum is white. The first maximum is a full spectrum with violet on the inside

and red on the outside. The second maximum is also a full spectrum, with red in it

overlapping with violet in the third maximum. At larger angles, the light soon starts

mixing to white again.

8.

(a) Two waves interfere constructively if their path difference is either zero or some

integral multiple of the wavelength; that is, if the path difference is mλ, where m is an

integer. (b) Two waves interfere destructively if their path difference is an odd multiple of

1

λ

one-half of a wavelength; that is, if the path difference equals m + λ = ( 2m + 1) .

2

2

10.

The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When

the finger touches a smooth surface, the oils from the skin will be deposited on the surface

in the pattern of the closely spaced ridges. The clear spaces between the lines of deposited

oil can serve as the slits in a crude diffraction grating and produce a colored spectrum of

the light passing through or reflecting from the glass surface.

12.

Suppose the index of refraction of the coating is intermediate between vacuum and the

glass. When the coating is very thin, light reflected from its top and bottom surfaces will

interfere constructively, so you see the surface white and brighter. Once the thickness

reaches one-quarter of the wavelength of violet light in the coating, destructive

interference for violet light will make the surface look red. Then other colors in spectral

order (blue, green, yellow, orange, and red) will interfere destructively, making the

surface look red, violet, and then blue. As the coating gets thicker, constructive

interference is observed for violet light and then for other colors in spectral order. Even

thicker coatings give constructive and destructive interference for several visible

wavelengths, so the reflected light starts looking white again.

14.

The reflected light is partially polarized, with the component parallel to the reflecting

surface being the most intense. Therefore, the polarizing material should have its

transmission axis oriented in the vertical direction in order to minimize the intensity of the

reflected light from horizontal surfaces.

Wave Optics

305

16.

One way to produce interference patterns is to allow light to pass through very small

openings. The opening between threads in a tautly stretched cloth like that in an umbrella

is small enough for the effects to be observed.

18.

Sound waves are longitudinal waves and cannot be polarized.

20.

The first experiment. The separation between maxima is inversely proportional to the slit

separation (see Eq. 24.5), so increasing the slit separation causes the distance between the

two maxima to decrease.

22.

The separations are greater in the second experiment when using red light having the

longer wavelength.

306

CHAPTER 24

Answers to Even Numbered Problems

1.77 µ m

2.

(a)

4.

2.61 m

6.

1.5 mm

8.

3.00 cm

1.93 µ m

(b)

1.47 µ m

(b)

δ = 3λ

(b)

81.58 nm

(b)

4.5 mm

10.

(a)

12.

1.73 km

14.

(a)

16.

193 nm

18.

233 nm

20.

290 nm

22.

8 (counting the zeroth order)

24.

6.5 × 10 2 nm

26.

99.6 nm

28.

20.0 × 10 −6 ( °C )

30.

(a)

32.

91.2 cm

34.

0.227 mm

36.

(a)

2 complete orders

(b)

10.9°

38.

(a)

13 orders

(b)

1 order

40.

7.35°

42.

469 nm and 78.1 nm

44.

632.8 nm

46.

38

123.4 nm

−1

2.3 mm

(c)

maximum

Wave Optics

48.

36.9°

50.

60.5°

52.

(a)

54.7°

(b)

63.4°

54.

(a)

I I0 = 1 2

(b)

54.7°

56.

432 nm

58.

maxima at 0°, 29.1°, and 76.3°

minima at 14.1° and 46.8°

60.

113 dark fringes

62.

(a)

(b)

0.25

68.

313 nm

70.

74 µ m

0

(c)

71.6°

307

308

CHAPTER 24

Problem Solutions

24.1

∆ybright = ym+1 − ym =

( 632.8 × 10

=

λL

−9

d

( m + 1) −

d

m ) ( 5.00 m )

0.200 × 10 −3 m

24.2

λL

m=

λL

d

= 1.58 × 10 −2 m = 1.58 cm

(a) For a bright fringe of order m, the path difference is δ = mλ , where m = 0, 1, 2,… At

the location of the third order bright fringe, m = 3 and

δ = 3 λ = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 µ m

1

(b) For a dark fringe, the path difference is δ = m + λ , where m = 0,1, 2, …

2

At the third dark fringe, m = 2 and

1

5

δ = 2 + λ = ( 589 nm ) = 1.47 × 10 3 nm = 1.47 µ m

24.3

2

2

(a) The distance between the central maximum and the first order bright fringe is

λL

∆y = ybright

− ybright

=

, or

m =1

m=0

d

∆y =

λL

d

=

( 546.1 × 10

−9

m ) ( 1.20 m )

0.250 × 10 −3 m

= 2.62 × 10 −3 m = 2.62 mm

(b) The distance between the first and second dark bands is

∆y = ydark

24.4

m =1

− ydark

m=0

=

λL

d

= 2.62 mm as in (a) above.

λL

1

m + , the spacing between the first and second dark fringes is

2

d

λL 3 1 λL

∆y =

. Thus, the required distance to the screen is

− =

d 2 2 d

From ydark =

L=

( ∆y ) d = ( 4.00 × 10−3 m )( 0.300 × 10-3 m ) =

λ

460 × 10 −9 m

2.61 m

Wave Optics

24.5

(a) From d sin θ = mλ , the angle for the m = 1 maximum for the sound waves is

m v

m

λ = sin −1 sound

d

d f

θ = sin −1

354 m s

1

−1

= sin

= 36.2°

0.300 m 2 000 Hz

(b) For 3.00-cm microwaves, the required slit spacing is

d=

( 1)( 3.00 cm )

mλ

=

= 5.08 cm

sin θ

sin36.2°

(c) The wavelength is λ =

d sin θ

; and if this is light, the frequency is

m

( 1) ( 3.00 × 108 m s )

mc

f= =

=

= 5.08 × 1014 Hz

λ d sin θ ( 1.00 × 10 -6 m ) sin 36.2°

c

24.6

The position of the first order bright fringe for wavelength λ is y1 =

λL

d

−9

( ∆λ ) L ( 700 − 400 ) × 10 m ( 1.5 m )

Thus, ∆y1 =

=

= 1.5 × 10 −3 m = 1.5 mm

−3

0.30 × 10

d

24.7

m

Note that, with the conditions given, the small angle

approximation does not work well. That is,

sinθ , tanθ , and θ are significantly different.

The approach to be used is outlined below.

(a) At the m = 2 maximum, δ = d sin θ = 2 λ ,

or λ =

or λ =

y

d

d

sin θ =

2

2

2 L + y2

( 300 m )

2

400 m

2

(1 000 m ) + ( 400 m )2

400 m

300 m

= 55.7 m

1 000 m

309

310

CHAPTER 24

(b) The next minimum encountered is the m = 2 minimum; and at that point,

1

5

δ = d sin θ = m + λ = λ

2

2

5λ

−1 5 ( 55.7 m )

or θ = sin −1

= 27.7°

= sin

2d

2 ( 300 m )

y = ( 1 000 m ) tan 27.7° = 524 m

Then,

so the car must travel an additional 124 m

24.8

In a double-slit interference pattern the distance from the central maximum to the

position of the mth order bright fringe is given by

λL

ym = m

d

where d is the distance between the splits and L is the distance to the screen. Thus, the

spacing between the first- and second-order bright fringes is

( 600 × 10 −9 m ) ( 2.50 m )

λL

= 0.0300 m = 3.00 cm

∆y = y2 − y1 = [2 − 1]

= 1

0.050 × 10 −3 m

d

24.9

The path difference in the two waves received at the home is δ = 2 d , where d is the

distance from the home to the mountain. Neglecting any phase change upon reflection,

the condition for destructive interference is

1

δ = m + λ with m = 0, 1, 2,…

so

24.10

dmin =

2

δ min

2

1 λ λ 300 m

= 0 + = =

= 75.0 m

2

4

2 4

The angular deviation from the line of the central maximum is given by

y

θ = tan −1 = tan −1

140 cm

L

1.80 cm

= 0.737°

311

Wave Optics

(a) The path difference is then

δ = d sin θ = ( 0.150 mm ) sin ( 0.737° ) = 1.93 × 10−3 mm = 1.93 µ m

(b)

λ

δ = ( 1.93 × 10 −6 m )

643 × 10

−9

= 3.00 λ

m

(c) Since the path difference for this position is a whole number of wavelengths, the

waves interfere constructively and produce a maximum at this spot.

24.11

The distance between the central maximum (position of A) and the first minimum is

λL

1

λL

y=

=

m+

d

2 m=0 2 d

Thus, d =

24.12

( 3.00 m )( 150 m )

= 11.3 m

2y

2 ( 20.0 m )

λL

=

The path difference in the two waves received

at the home is δ = 2d − 30.0 km where d is defined

in the figure at the right. For minimum cloud

height and (hence minimum path difference) to

yield destructive interference, δ = λ 2 giving

1

λ

dmin = 30.0 km + = 15.1 km , and

2

2

2

hmin = dmin

− ( 15.0 km ) =

2

24.13

Cloud

d

d

h

15.0 km

Transmitter

15.0 km

Receiver

2

2

(15.1 km ) − ( 15.0 km ) = 1.73 km

As shown in the figure at the right, the path

difference in the waves reaching the telescope is

δ = d2 − d1 = d2 ( 1 − sin α ) . If the first minimum

(δ = λ 2 )

d1

occurs when θ = 25.0° , then

α = 180° − (θ + 90.0° + θ ) = 40.0° , and

d2 =

δ

1 − sin α

=

( 250 m 2 )

1 − sin 40.0°

Thus, h = d2 sin 25.0° = 148 m

q

= 350 m

a

q

d2

h

312

CHAPTER 24

24.14

With n film > nair , light reflecting from the front surface of the film (an air to film

boundary) experiences a 180° phase shift, but light reflecting from the back surface (a

film to air boundary) experiences no shift. The condition for constructive interference in

the two reflected waves is then

1

2n filmt = m + λ

2

with

For minimum thickness, m = 0 , giving

24.15

m = 0, 1, 2, …

tmin =

λ

4 n film

(a)

With λ = 656.3 nm and n film = 1.330 , tmin =

656.3 nm

= 123.4 nm

4 ( 1.330 )

(b)

When λ = 434.0 nm and n film = 1.330 , tmin =

434.0 nm

= 81.58 nm

4 ( 1.330 )

Light reflecting from the upper surface undergoes phase reversal while that reflecting

from the lower surface does not. The condition for constructive interference in the

reflected light is then

2t −

λn

1λ

1 λ

, m = 0, 1, 2, …

= mλn , or t = m + n = m +

2

2 2

2 2 n film

For minimum thickness, m = 0 giving

t=

24.16

λ

4 n film

=

500 nm

= 91.9 nm

4 ( 1.36 )

With nglass > nair and nliquid < nglass , light reflecting from the air-glass boundary experiences

a 180° phase shift, but light reflecting from the glass-liquid boundary experiences no

shift. Thus, the condition for destructive interference in the two reflected waves is

2nglass t = mλ

where

m = 0, 1, 2, …

For minimum (non-zero) thickness, m = 1 giving

t=

λ

2nglass

=

580 nm

= 193 nm

2 ( 1.50 )

Wave Optics

24.17

313

With ncoating > nair and ncoating > nlens , light reflecting at the air-coating boundary

experiences a phase reversal, but light reflecting from the coating-lens boundary does

not. Therefore, the condition for destructive interference in the two reflected waves is

2ncoating t = mλ

where

m = 0, 1, 2,…

For finite wavelengths, the lowest allowed value of m is m = 1 . Then, if

t = 177.4 nm and ncoating = 1.55 , the wavelength associated with this lowest order

destructive interference is

λ1 =

24.18

2ncoating t

= 2 ( 1.55 )( 177.4 nm ) = 550 nm

1

Since nair < noil < nwater , light reflected from both top and bottom surfaces of the oil film

experiences phase reversal, resulting in zero net phase difference due to reflections.

Therefore, the condition for constructive interference in reflected light is

2 t = mλn = m

λ

n film

λ

, or t = m

2 n film

where m = 0, 1, 2,…

Assuming that m = 1 , the thickness of the oil slick is

t = (1)

24.19

λ

2 n film

=

600 nm

= 233 nm

2 ( 1.29 )

There will be a phase reversal of the radar waves reflecting from both surfaces of the

polymer, giving zero net phase change due to reflections. The requirement for

destructive interference in the reflected waves is then

1

λ

where m = 0, 1, 2,…

2 t = m + λn , or t = ( 2m + 1)

4 n film

2

If the film is as thin as possible, then m = 0 and the needed thickness is

t=

λ

4 n film

=

3.00 cm

= 0.500 cm

4 ( 1.50 )

This anti-reflectance coating could be easily countered by changing the wavelength of

the radar—to 1.50 cm—now creating maximum reflection!

314

CHAPTER 24

24.20

The transmitted light is brightest when the reflected light is a minimum (that is, the

same conditions that produce destructive interference in the reflected light will produce

constructive interference in the transmitted light). As light enters the air layer from

glass, any light reflected at this surface has zero phase change. Light reflected from the

other surface of the air layer (where light is going from air into glass) does have a phase

reversal. Thus, the condition for destructive interference in the light reflected from the

air film is 2 t = m λn , m = 0, 1, 2,…

Since λn =

λ

n film

condition is

24.21

=

λ

1.00

= λ , the minimum non-zero plate separation satisfying this

d = t = ( 1)

λ

2

=

580 nm

= 290 nm

2

(a) For maximum transmission, we want destructive interference in the light reflected

from the front and back surfaces of the film.

If the surrounding glass has refractive index greater than 1.378, light reflected from

the front surface suffers no phase reversal, and light reflected from the back does

undergo phase reversal. This effect by itself would produce destructive interference,

so we want the distance down and back to be one whole wavelength in the film.

Thus, we require that

2t = λn = λ n film or

t=

λ

2 n film

=

656.3 nm

= 238 nm

2 ( 1.378 )

(b) The filter will expand. As t increases in 2n film t = λ , so does λ increase

(c) Destructive interference for reflected light happens also for λ in 2 t = 2 λ n film , or

λ = n filmt = ( 1.378 )( 238 nm ) = 328 nm (near ultraviolet)

Wave Optics

24.22

315

Light reflecting from the lower surface of the air layer experiences phase reversal, but

light reflecting from the upper surface of the layer does not. The requirement for a dark

fringe (destructive interference) is then

λ

2 t = mλn = m

= m λ , where m = 0, 1, 2,…

nair

At the thickest part of the film ( t = 2.00 µ m ) , the order number is

m=

2t

λ

=

2 ( 2.00 × 10 −6 m )

546.1 × 10 −9 m

= 7.32

Since m must be an integer, m = 7 is the order of the last dark fringe seen. Counting the

m = 0 order along the edge of contact, a total of 8 dark fringes will be seen.

24.23

With a phase reversal upon reflection from the

lower surface of the air layer and no phase change

for reflection at the upper surface of the layer, the

condition for destructive interference is

λ

2 t = mλn = m

= m λ , where m = 0, 1, 2,…

nair

Counting the zeroth order along the edge of

contact, the order number of the thirtieth dark

fringe observed is m = 29 . The thickness of the air

layer at this point is t = 2r , where r is the radius of

the wire. Thus,

−9

t 29 λ 29 ( 600 × 10 m )

=

= 4.35 µ m

r= =

2

4

4

Incident

light

n

t3

t2

t1

O

316

CHAPTER 24

h=Rt

24.24

R

Glass

air

no phase reversal

phase reversal

r

t{

Glass

From the geometry shown in the figure, R2 = ( R − t ) + r 2 , or

2

t = R − R2 − r 2

= 3.0 m −

2

( 3.0 m ) − ( 9.8 × 10 −3 m )

2

= 1.6 × 10 −5 m

With a phase reversal upon reflection at the lower surface of the air layer, but no

reversal with reflection from the upper surface, the condition for a bright fringe is

1

2 t = m + λn = m +

2

1 λ

= m+

2 nair

1

λ , where m = 0, 1, 2,…

2

At the 50 th bright fringe, m = 49 , and the wavelength is found to be

2 ( 1.6 × 10 −5 m )

2t

λ=

=

= 6.5 × 10 −7 m = 6.5 × 10 2 nm

m+1 2

49.5

24.25

There is a phase reversal due to reflection at the bottom of the air film but not at the top

of the film. The requirement for a dark fringe is then

2 t = mλn = m

λ

nair

= mλ , where m = 0, 1, 2,…

At the 19th dark ring (in addition to the dark center spot), the order number is m = 19 ,

and the thickness of the film is

mλ 19 ( 500 × 10

=

t=

2

2

−9

m)

= 4.75 × 10 −6 m = 4.75 µ m

Wave Optics

24.26

317

With a phase reversal due to reflection at each surface of the magnesium fluoride layer,

there is zero net phase difference caused by reflections. The condition for destructive

interference is then

2t = m +

1

λn = m +

2

1 λ

, where m = 0, 1, 2,…

2 n film

For minimum thickness, m = 0 , and the thickness is

t = ( 2 m + 1)

24.27

λ

4 n film

( 550 × 10

= ( 1)

−9

4 ( 1.38 )

m)

= 9.96 × 10 −8 m = 99.6 nm

There is a phase reversal upon reflection at each surface of the film and hence zero net

phase difference due to reflections. The requirement for constructive interference in the

reflected light is then

2 t = mλn = m

λ

n film

, where m = 1, 2, 3, …

With t = 1.00 × 10 −5 cm = 100 nm , and n film = 1.38 , the wavelengths intensified in the

reflected light are

λ=

2 n film t

m

=

2 ( 1.38 )( 100 nm )

, with m = 1, 2, 3, …

m

Thus, λ = 276 nm, 138 nm, 92.0 nm …

and none of these wavelengths are in the visible spectrum

318

CHAPTER 24

24.28

As light emerging from the glass reflects from the top of the air layer, there is no phase

reversal produced. However, the light reflecting from the end of the metal rod at the

bottom of the air layer does experience phase reversal. Thus, the condition for

constructive interference in the reflected light is 2 t = ( m + 12 ) λair .

As the metal rod expands, the thickness of the air layer decreases. The increase in the

length of the rod is given by

∆L = ∆t = ( mi +

1

2

)

λair

2

(

− mf +

1

2

) λ2

air

= ∆m

λair

2

The order number changes by one each time the film changes from bright to dark and

back to bright. Thus, during the expansion, the measured change in the length of the rod

is

∆L = ( 200 )

λair

2

= ( 200 )

( 500 × 10

2

−9

m)

= 5.00 × 10 −5 m

From ∆L = L0α ( ∆T ) , the coefficient of linear expansion of the rod is

α=

24.29

∆L

5.00 × 10 −5 m

−1

=

= 20.0 × 10 −6 ( °C )

L0 ( ∆T ) ( 0.100 m ) ( 25.0 °C )

The distance on the screen from the center to either edge of the central maximum is

λ

y = L tan θ ≈ L sin θ = L

a

632.8 × 10 −9 m

−3

= ( 1.00 m )

= 2.11 × 10 m=2.11 mm

−3

0.300 × 10 m

The full width of the central maximum on the screen is then

2 y = 4.22 mm

Wave Optics

24.30

319

(a) Dark bands occur where sin θ = m ( λ a ) . At the first dark band, m = 1 , and the

distance from the center of the central maximum is

λ

y1 = L tan θ ≈ L sin θ = L

a

600 × 10 −9 m

−3

= ( 1.5 m )

= 2.25 × 10 m = 2.3 mm

−3

0.40 × 10 m

(b) The width of the central maximum is 2 y1 = 2 ( 2.25 mm ) = 4.5 mm

24.31

(a) Dark bands (minima) occur where sin θ = m ( λ a ) . For the first minimum, m = 1 and

the distance from the center of the central maximum is y1 = L tan θ ≈ L sin θ = L ( λ a ) .

Thus, the needed distance to the screen is

0.75 × 10 −3 m

a

L = y1 = ( 0.85 × 10 −3 m )

= 1.1 m

-9

λ

587.5 × 10 m

(b) The width of the central maximum is 2 y1 = 2 ( 0.85 mm ) = 1.7 mm

Note: The small angle approximation does not

work well in this situation. Rather, you should

proceed as follows.

At the first order minimum, sin θ = λ a or

λ

5.00 cm

θ = sin −1 = sin −1

= 7.98°

a

36.0 cm

a = 36.0 cm = 0.360 m

L = 6.50 m

24.32

Central

Maximum

Then, y1 = L tan θ = ( 6.50 m ) tan 7.98° = 0.912 m = 91.2 cm

q

y1 First Order

Minimum

320

CHAPTER 24

24.33

The locations of the dark fringes (minima) mark the edges of the maxima, and the

widths of the maxima equals the spacing between successive minima.

At the locations of the minima, sin θ m = m ( λ a ) and

λ

ym = L tan θ m ≈ L sin θ m = m L

a

500 × 10 −9 m

= m ( 1.20 m )

= m ( 1.20 mm )

-3

0.500 × 10 m

Then, ∆y = ∆m ( 1.20 mm ) and for successive minima, ∆m = 1 .

Therefore, the width of each maxima, other than the central maximum, in this interference

pattern is

width = ∆y = ( 1)( 1.20 mm ) = 1.20 mm

24.34

At the positions of the minima, sin θ m = m ( λ a ) and

ym = L tan θ m ≈ L sin θ m = m L ( λ a )

Thus, y3 − y1 = ( 3 − 1) L ( λ a ) = 2 L ( λ a )

and

24.35

2 ( 0.500 m ) ( 680 × 10 −9 m )

2Lλ

a=

=

= 2.27 × 10 −4 m = 0.227 mm

−3

y3 − y1

3.00 × 10 m

The grating spacing is d =

1

1

cm =

m and d sin θ = mλ

3 660

3.66 × 10 5

(a) The wavelength observed in the first-order spectrum is λ = d sin θ , or

10 4 nm

1 m 109 nm

sin

θ

=

sin θ

5

3.66 × 10 1 m

3.66

λ =

This yields:

at 10.1°, λ = 479 nm ;

and

at 14.8°, λ = 698 nm

at 13.7°, λ = 647 nm ;

Wave Optics

321

(b) In the second order, m = 2 . The second order images for the above wavelengths will

be found at angles θ 2 = sin −1 ( 2 λ d ) = sin −1 [ 2sin θ1 ]

24.36

This yields:

for λ = 479 nm , θ 2 = 20.5° ;

and

for λ = 698 nm , θ 2 = 30.7°

for λ = 647 nm , θ 2 = 28.3° ;

(a) The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is

1 mm

d=

= 1.67 × 10 −3 mm = 1.67 × 10 −6 m

600

Thus, mmax =

d sin 90.0°

λred

(1.67 × 10

=

−6

m ) sin 90.0°

700 × 10 −9 m

= 2.38

so 2 complete orders will be observed.

(b) From λ = d sin θ , the angular separation of the red and violet edges in the first order

will be

−9

700 × 10 −9 m

m

λ

λ

−1 400 × 10

∆θ = sin −1 red − sin −1 violet = sin −1

−

sin

−6

−6

d

d

1.67 × 10 m

1.67 × 10 m

or ∆θ = 10.9°

24.37

1 cm

1m

=

. From d sin θ = mλ , the angular separation

4 500 4.50 × 10 5

between the given spectral lines will be

The grating spacing is d =

m λred

m λviolet

∆θ = sin −1

− sin −1

d

d

or

m ( 656 × 10 −9 m )( 4.50 × 10 5 )

m ( 434 × 10 −9 m )( 4.50 × 10 5 )

− sin −1

∆θ = sin −1

1m

1m

The results obtained are: for m = 1, ∆θ = 5.91° ; for m = 2, ∆θ = 13.2° ;

and for m = 3, ∆θ = 26.5° . Complete orders for m ≥ 4 are not visible.

322

CHAPTER 24

24.38

(a) If d =

1 cm

= 6.67 × 10 −4 cm = 6.67 × 10 −6 m , the highest order of λ = 500 nm that can

1 500

be observed will be

mmax =

(b) If d =

λ

=

( 6.67 × 10

−6

500 × 10

m ) ( 1)

−9

m

= 13.3 or 13 orders

1 cm

= 6.67 × 10 −5 cm = 6.67 × 10 −7 m , then

15 000

mmax =

24.39

d sin 90°

d sin 90°

λ

( 6.67 × 10

=

−7

m ) ( 1)

500 × 10 −9 m

= 1.33 or 1 order

1 cm

= 2.00 × 10 −4 cm = 2.00 × 10 −6 m , and d sin θ = mλ gives

5 000

the angular position of a second order spectral line as

The grating spacing is d =

sin θ =

2λ

2λ

or θ = sin −1

d

d

For the given wavelengths, the angular positions are

2 ( 610 × 10 −9 m )

2 ( 480 × 10 −9 m )

−1

= 28.7°

θ1 = sin

= 37.6° and θ 2 = sin

−6

−6

2.00 × 10 m

2.00 × 10 m

−1

If L is the distance from the grating to the screen, the distance on the screen from the

central maximum to a second order bright line is y = L tan θ . Therefore, for the two

given wavelengths, the screen separation is

∆y = L [ tan θ1 − tan θ 2 ]

= ( 2.00 m ) tan ( 37.6° ) − tan ( 28.7° ) = 0.445 m = 44.5 cm

323

Wave Optics

24.40

With 2 000 lines per centimeter, the grating spacing is

d=

1

cm = 5.00 × 10 −4 cm = 5.00 × 10 −6 m

2 000

Then, from d sin θ = mλ , the location of the first order for the red light is

( 1) ( 640 × 10 −9 m )

mλ

−1

= 7.35°

θ = sin

= sin

−6

d

5.00 × 10 m

−1

24.41

The grating spacing is d =

1 cm 10 −2 m

=

= 3.636 × 10 −6 m . From d sin θ = mλ , or

2 750

2 750

θ = sin −1 ( mλ d ) , the angular positions of the red and violet edges of the second-order

spectrum are found to be

2 ( 700 × 10 −9 m )

2λred

−1

= 22.65°

θ r = sin

= sin

3.636 × 10 −6 m

d

−1

and

2λviolet

d

θ v = sin −1

2 ( 400 × 10 −9 m )

−1

= 12.71°

sin

=

3.636 × 10 −6 m

Note from the sketch at the right that yr = L tan θ r

and yv = L tan θ v , so the width of the spectrum on

Grating

Screen

Dy

the screen is ∆y = L ( tan θ r − tan θ v ) .

Since it is given that ∆y = 1.75 cm , the distance

from the grating to the screen must be

L=

or

∆y

1.75 cm

=

tan θ r − tan θ v tan ( 22.65° ) − tan ( 12.71° )

L = 9.13 cm

qv

L

qr

yv

yr

324

CHAPTER 24

24.42

The grating spacing is d =

1 cm

= 8.33 × 10 −4 cm = 8.33 × 10 −6 m

1 200

mλ

and the small angle approximation, the distance from the central

d

maximum to the maximum of order m for wavelength λ is

ym = L tan θ ≈ L sin θ = ( λ L d ) m . Therefore, the spacing between successive maxima is

Using sin θ =

∆y = ym+1 − ym = λ L d .

The longer wavelength in the light is found to be

λlong =

( ∆y ) d = ( 8.44 × 10−3 m )( 8.33 × 10−6 m ) =

0.150 m

L

469 nm

Since the third order maximum of the shorter wavelength falls halfway between the

central maximum and the first order maximum of the longer wavelength, we have

3 λshort L 0 + 1 λlong L

1

or λshort = ( 469 nm ) = 78.1 nm

=

d

2 d

6

24.43

The grating spacing is d =

1 mm

= 2.50 × 10 −3 mm = 2.50 × 10 −6 m

400

From d sin θ = mλ , the angle of the second-order diffracted ray is θ = sin −1 ( 2λ d ) .

(a) When the grating is surrounded by air, the wavelength is λair = λ nair ≈ λ and

2 ( 541 × 10 −9 m )

2 λair

−1

= 25.6°

θ a = sin

= sin

−6

d

2.50 × 10 m

−1

(b) If the grating is immersed in water,

then λn = λwater =

λ

nwater

2 λwater

d

θ b = sin −1

=

λ

1.333

, yielding

2 ( 541 × 10 −9 m )

−1

= 19.0°

sin

=

−6

( 2.50 × 10 m ) ( 1.333 )

mλn m ( λ n )

mλ

=

, we have that n sin θ =

= constant when m is kept

d

d

d

constant. Therefore nair sin θ a = nwater sin θ b , or the angles of parts (a) and (b) satisfy

(c) From sin θ =

Snell’s law.

Wave Optics

24.44

When light of wavelength λ passes through a single slit of width a, the first minimum is

observed at angle θ where

sin θ =

λ

λ

or θ = sin −1

a

a

This will have no solution if a < λ , so the maximum slit width if no minima are to be

seen is a = λ = 632 .8 nm .

24.45

(a) From Brewster’s law, the index of refraction is

n2 = tan θ p = tan ( 48.0° ) = 1.11

(b) From Snell’s law, n2 sin θ 2 = n1 sin θ1 , we obtain when θ1 = θ p

n1 sin θ p

−1 ( 1.00 ) sin 48.0°

= sin

= 42.0°

1.11

n2

θ 2 = sin −1

Note that when θ1 = θ p , θ 2 = 90.0° − θ p as it should.

24.46

Unpolarized light incident on a polarizer contains electric field vectors at all angles to

the transmission axis of the polarizer. Malus’s law then gives the intensity of the

transmitted light as I = I 0 ( cos 2 θ )av . Since the average value of cos 2 θ is 1 2 , the

intensity of the light passed by the first polarizer is I1 = I 0 2 , where I 0 is the incident

intensity.

Then, from Malus’s law, the intensity passed by the second polarizer is

I

3

I 3

I 2 = I1 cos 2 ( 30.0° ) = 0 , or 2 =

I0

8

2 4

24.47

325

The more general expression for Brewster’s angle is (see Problem 51)

tan θ p = n2 n1

n

1.52

(a) When n1 = 1.00 and n2 = 1.52 , θ p = tan −1 2 = tan −1

= 56.7°

1.00

n1

(b)

n

When n1 = 1.333 and n2 = 1.52 , θ p = tan −1 2

n1

−1 1.52

= tan

= 48.8°

1.333

326

CHAPTER 24

24.48

The polarizing angle for light in air striking a water surface is

n2

n1

θ p = tan −1

−1 1.333

= tan

= 53.1°

1.00

This is the angle of incidence for the incoming sunlight (that is, the angle between the

incident light and the normal to the surface). The altitude of the Sun is the angle

between the incident light and the water surface. Thus, the altitude of the Sun is

α = 90.0° − θ p = 90.0° − 53.1° = 36.9°

24.49

n

1.65

The polarizing angle is θ p = tan −1 2 = tan −1

= 58.8°

1.00

n1

When the light is incident at the polarizing angle, the angle of refraction is

θ r = 90.0° − θ p = 90.0° − 58.8° = 31.2°

24.50

The critical angle for total reflection is θ c = sin −1 ( n2 n1 ) . Thus, if θ c = 34.4° as light

attempts to go from sapphire into air, the index of refraction of sapphire is

nsapphire = n1 =

n2

1.00

=

= 1.77

sin θ c sin 34.4°

Then, when light is incident on sapphire from air, the Brewster angle is

n2

−1 1.77

= tan

= 60.5°

1.00

n1

θ p = tan −1

24.51

From Snell’s law, the angles of incidence and refraction are related by n1 sin θ1 = n2 sin θ 2 .

If the angle of incidence is the polarizing angle (that is, θ1 = θ p ), the angles of incidence

and refraction are also related by

θ p + θ 2 + 90° = 180° , or θ 2 = 90° − θ p

Substitution into Snell’s law then gives

(

)

n1 sin θ p = n2 sin 90° − θ p = n2 cosθ p or tan θ p = n2 n1

Wave Optics

24.52

24.53

I = I 0 cos 2 θ

I

I0

⇒

θ = cos −1

(a)

I

1

=

⇒

I 0 3.00

θ = cos −1

(b)

I

1

=

⇒

I 0 5.00

θ = cos −1

(c)

I

1

=

I 0 10.0

θ = cos −1

⇒

1

= 54.7°

3.00

1

= 63.4°

5.00

1

= 71.6°

10.0

From Malus’s law, the intensity of the light transmitted by the first polarizer is

I1 = I i cos 2 θ1 . The plane of polarization of this light is parallel to the axis of the first plate

and is incident on the second plate. Malus’s law gives the intensity transmitted by the

second plate as I 2 = I1 cos 2 (θ 2 − θ1 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) . This light is polarized parallel

to the axis of the second plate and is incident upon the third plate. A final application of

Malus’s law gives the transmitted intensity as

I f = I 2 cos 2 (θ 3 − θ 2 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) cos 2 (θ 3 − θ 2 )

With θ1 = 20.0°, θ 2 = 40.0°, and θ 3 = 60.0° , this result yields

I f = ( 10.0 units ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) = 6.89 units

24.54

327

(a) Using Malus’s law, the intensity of the transmitted light is found to be

(

I = I 0 cos 2 ( 45° ) = I 0 1

)

2

2 , or I I 0 = 1 2

(b) From Malus’s law, I I 0 = cos 2 θ . Thus, if I I 0 = 1 3 we obtain

(

cos 2 θ = 1 3 or θ = cos −1 1

)

3 = 54.7°

Wave Optics

Quick Quizzes

1.

(c). The fringes on the screen are equally spaced only at small angles where tan θ ≈ sin θ is

a valid approximation.

2.

(b). The space between successive bright fringes is proportional to the wavelength of the

light. Since the wavelength in water is less than that in air, the bright fringes are closer

together in the second experiment.

3.

(b). The outer edges of the central maximum occur where sin θ = ± λ a . Thus, as a, the

width of the slit, becomes smaller, the width of the central maximum will increase.

4.

The compact disc. The tracks of information on a compact disc are much closer together

than on a phonograph record. As a result, the diffraction maxima from the compact disc

will be farther apart than those from the record.

303

304

CHAPTER 24

Answers to Even Numbered Conceptual Questions

2.

The wavelength of light is extremely small in comparison to the dimensions of your hand,

so the diffraction of light around obstacles the size of your hand is totally negligible.

However, sound waves have wavelengths that are comparable to the dimensions of the

hand or even larger. Therefore, significant diffraction of sound waves occurs around hand

sized obstacles.

4.

The wavelength of light traveling in water would decrease, since the wavelength of light

in a medium is given by λn = λ n , where λ is the wavelength in vacuum and n is the

index of refraction of the medium. Since the positions of the bright and dark fringes are

proportional to the wavelength, the fringe separations would decrease.

6.

Every color produces its own interference pattern, and we see them superimposed. The

central maximum is white. The first maximum is a full spectrum with violet on the inside

and red on the outside. The second maximum is also a full spectrum, with red in it

overlapping with violet in the third maximum. At larger angles, the light soon starts

mixing to white again.

8.

(a) Two waves interfere constructively if their path difference is either zero or some

integral multiple of the wavelength; that is, if the path difference is mλ, where m is an

integer. (b) Two waves interfere destructively if their path difference is an odd multiple of

1

λ

one-half of a wavelength; that is, if the path difference equals m + λ = ( 2m + 1) .

2

2

10.

The skin on the tip of a finger has a series of closely spaced ridges and swirls on it. When

the finger touches a smooth surface, the oils from the skin will be deposited on the surface

in the pattern of the closely spaced ridges. The clear spaces between the lines of deposited

oil can serve as the slits in a crude diffraction grating and produce a colored spectrum of

the light passing through or reflecting from the glass surface.

12.

Suppose the index of refraction of the coating is intermediate between vacuum and the

glass. When the coating is very thin, light reflected from its top and bottom surfaces will

interfere constructively, so you see the surface white and brighter. Once the thickness

reaches one-quarter of the wavelength of violet light in the coating, destructive

interference for violet light will make the surface look red. Then other colors in spectral

order (blue, green, yellow, orange, and red) will interfere destructively, making the

surface look red, violet, and then blue. As the coating gets thicker, constructive

interference is observed for violet light and then for other colors in spectral order. Even

thicker coatings give constructive and destructive interference for several visible

wavelengths, so the reflected light starts looking white again.

14.

The reflected light is partially polarized, with the component parallel to the reflecting

surface being the most intense. Therefore, the polarizing material should have its

transmission axis oriented in the vertical direction in order to minimize the intensity of the

reflected light from horizontal surfaces.

Wave Optics

305

16.

One way to produce interference patterns is to allow light to pass through very small

openings. The opening between threads in a tautly stretched cloth like that in an umbrella

is small enough for the effects to be observed.

18.

Sound waves are longitudinal waves and cannot be polarized.

20.

The first experiment. The separation between maxima is inversely proportional to the slit

separation (see Eq. 24.5), so increasing the slit separation causes the distance between the

two maxima to decrease.

22.

The separations are greater in the second experiment when using red light having the

longer wavelength.

306

CHAPTER 24

Answers to Even Numbered Problems

1.77 µ m

2.

(a)

4.

2.61 m

6.

1.5 mm

8.

3.00 cm

1.93 µ m

(b)

1.47 µ m

(b)

δ = 3λ

(b)

81.58 nm

(b)

4.5 mm

10.

(a)

12.

1.73 km

14.

(a)

16.

193 nm

18.

233 nm

20.

290 nm

22.

8 (counting the zeroth order)

24.

6.5 × 10 2 nm

26.

99.6 nm

28.

20.0 × 10 −6 ( °C )

30.

(a)

32.

91.2 cm

34.

0.227 mm

36.

(a)

2 complete orders

(b)

10.9°

38.

(a)

13 orders

(b)

1 order

40.

7.35°

42.

469 nm and 78.1 nm

44.

632.8 nm

46.

38

123.4 nm

−1

2.3 mm

(c)

maximum

Wave Optics

48.

36.9°

50.

60.5°

52.

(a)

54.7°

(b)

63.4°

54.

(a)

I I0 = 1 2

(b)

54.7°

56.

432 nm

58.

maxima at 0°, 29.1°, and 76.3°

minima at 14.1° and 46.8°

60.

113 dark fringes

62.

(a)

(b)

0.25

68.

313 nm

70.

74 µ m

0

(c)

71.6°

307

308

CHAPTER 24

Problem Solutions

24.1

∆ybright = ym+1 − ym =

( 632.8 × 10

=

λL

−9

d

( m + 1) −

d

m ) ( 5.00 m )

0.200 × 10 −3 m

24.2

λL

m=

λL

d

= 1.58 × 10 −2 m = 1.58 cm

(a) For a bright fringe of order m, the path difference is δ = mλ , where m = 0, 1, 2,… At

the location of the third order bright fringe, m = 3 and

δ = 3 λ = 3 ( 589 nm ) = 1.77 × 10 3 nm = 1.77 µ m

1

(b) For a dark fringe, the path difference is δ = m + λ , where m = 0,1, 2, …

2

At the third dark fringe, m = 2 and

1

5

δ = 2 + λ = ( 589 nm ) = 1.47 × 10 3 nm = 1.47 µ m

24.3

2

2

(a) The distance between the central maximum and the first order bright fringe is

λL

∆y = ybright

− ybright

=

, or

m =1

m=0

d

∆y =

λL

d

=

( 546.1 × 10

−9

m ) ( 1.20 m )

0.250 × 10 −3 m

= 2.62 × 10 −3 m = 2.62 mm

(b) The distance between the first and second dark bands is

∆y = ydark

24.4

m =1

− ydark

m=0

=

λL

d

= 2.62 mm as in (a) above.

λL

1

m + , the spacing between the first and second dark fringes is

2

d

λL 3 1 λL

∆y =

. Thus, the required distance to the screen is

− =

d 2 2 d

From ydark =

L=

( ∆y ) d = ( 4.00 × 10−3 m )( 0.300 × 10-3 m ) =

λ

460 × 10 −9 m

2.61 m

Wave Optics

24.5

(a) From d sin θ = mλ , the angle for the m = 1 maximum for the sound waves is

m v

m

λ = sin −1 sound

d

d f

θ = sin −1

354 m s

1

−1

= sin

= 36.2°

0.300 m 2 000 Hz

(b) For 3.00-cm microwaves, the required slit spacing is

d=

( 1)( 3.00 cm )

mλ

=

= 5.08 cm

sin θ

sin36.2°

(c) The wavelength is λ =

d sin θ

; and if this is light, the frequency is

m

( 1) ( 3.00 × 108 m s )

mc

f= =

=

= 5.08 × 1014 Hz

λ d sin θ ( 1.00 × 10 -6 m ) sin 36.2°

c

24.6

The position of the first order bright fringe for wavelength λ is y1 =

λL

d

−9

( ∆λ ) L ( 700 − 400 ) × 10 m ( 1.5 m )

Thus, ∆y1 =

=

= 1.5 × 10 −3 m = 1.5 mm

−3

0.30 × 10

d

24.7

m

Note that, with the conditions given, the small angle

approximation does not work well. That is,

sinθ , tanθ , and θ are significantly different.

The approach to be used is outlined below.

(a) At the m = 2 maximum, δ = d sin θ = 2 λ ,

or λ =

or λ =

y

d

d

sin θ =

2

2

2 L + y2

( 300 m )

2

400 m

2

(1 000 m ) + ( 400 m )2

400 m

300 m

= 55.7 m

1 000 m

309

310

CHAPTER 24

(b) The next minimum encountered is the m = 2 minimum; and at that point,

1

5

δ = d sin θ = m + λ = λ

2

2

5λ

−1 5 ( 55.7 m )

or θ = sin −1

= 27.7°

= sin

2d

2 ( 300 m )

y = ( 1 000 m ) tan 27.7° = 524 m

Then,

so the car must travel an additional 124 m

24.8

In a double-slit interference pattern the distance from the central maximum to the

position of the mth order bright fringe is given by

λL

ym = m

d

where d is the distance between the splits and L is the distance to the screen. Thus, the

spacing between the first- and second-order bright fringes is

( 600 × 10 −9 m ) ( 2.50 m )

λL

= 0.0300 m = 3.00 cm

∆y = y2 − y1 = [2 − 1]

= 1

0.050 × 10 −3 m

d

24.9

The path difference in the two waves received at the home is δ = 2 d , where d is the

distance from the home to the mountain. Neglecting any phase change upon reflection,

the condition for destructive interference is

1

δ = m + λ with m = 0, 1, 2,…

so

24.10

dmin =

2

δ min

2

1 λ λ 300 m

= 0 + = =

= 75.0 m

2

4

2 4

The angular deviation from the line of the central maximum is given by

y

θ = tan −1 = tan −1

140 cm

L

1.80 cm

= 0.737°

311

Wave Optics

(a) The path difference is then

δ = d sin θ = ( 0.150 mm ) sin ( 0.737° ) = 1.93 × 10−3 mm = 1.93 µ m

(b)

λ

δ = ( 1.93 × 10 −6 m )

643 × 10

−9

= 3.00 λ

m

(c) Since the path difference for this position is a whole number of wavelengths, the

waves interfere constructively and produce a maximum at this spot.

24.11

The distance between the central maximum (position of A) and the first minimum is

λL

1

λL

y=

=

m+

d

2 m=0 2 d

Thus, d =

24.12

( 3.00 m )( 150 m )

= 11.3 m

2y

2 ( 20.0 m )

λL

=

The path difference in the two waves received

at the home is δ = 2d − 30.0 km where d is defined

in the figure at the right. For minimum cloud

height and (hence minimum path difference) to

yield destructive interference, δ = λ 2 giving

1

λ

dmin = 30.0 km + = 15.1 km , and

2

2

2

hmin = dmin

− ( 15.0 km ) =

2

24.13

Cloud

d

d

h

15.0 km

Transmitter

15.0 km

Receiver

2

2

(15.1 km ) − ( 15.0 km ) = 1.73 km

As shown in the figure at the right, the path

difference in the waves reaching the telescope is

δ = d2 − d1 = d2 ( 1 − sin α ) . If the first minimum

(δ = λ 2 )

d1

occurs when θ = 25.0° , then

α = 180° − (θ + 90.0° + θ ) = 40.0° , and

d2 =

δ

1 − sin α

=

( 250 m 2 )

1 − sin 40.0°

Thus, h = d2 sin 25.0° = 148 m

q

= 350 m

a

q

d2

h

312

CHAPTER 24

24.14

With n film > nair , light reflecting from the front surface of the film (an air to film

boundary) experiences a 180° phase shift, but light reflecting from the back surface (a

film to air boundary) experiences no shift. The condition for constructive interference in

the two reflected waves is then

1

2n filmt = m + λ

2

with

For minimum thickness, m = 0 , giving

24.15

m = 0, 1, 2, …

tmin =

λ

4 n film

(a)

With λ = 656.3 nm and n film = 1.330 , tmin =

656.3 nm

= 123.4 nm

4 ( 1.330 )

(b)

When λ = 434.0 nm and n film = 1.330 , tmin =

434.0 nm

= 81.58 nm

4 ( 1.330 )

Light reflecting from the upper surface undergoes phase reversal while that reflecting

from the lower surface does not. The condition for constructive interference in the

reflected light is then

2t −

λn

1λ

1 λ

, m = 0, 1, 2, …

= mλn , or t = m + n = m +

2

2 2

2 2 n film

For minimum thickness, m = 0 giving

t=

24.16

λ

4 n film

=

500 nm

= 91.9 nm

4 ( 1.36 )

With nglass > nair and nliquid < nglass , light reflecting from the air-glass boundary experiences

a 180° phase shift, but light reflecting from the glass-liquid boundary experiences no

shift. Thus, the condition for destructive interference in the two reflected waves is

2nglass t = mλ

where

m = 0, 1, 2, …

For minimum (non-zero) thickness, m = 1 giving

t=

λ

2nglass

=

580 nm

= 193 nm

2 ( 1.50 )

Wave Optics

24.17

313

With ncoating > nair and ncoating > nlens , light reflecting at the air-coating boundary

experiences a phase reversal, but light reflecting from the coating-lens boundary does

not. Therefore, the condition for destructive interference in the two reflected waves is

2ncoating t = mλ

where

m = 0, 1, 2,…

For finite wavelengths, the lowest allowed value of m is m = 1 . Then, if

t = 177.4 nm and ncoating = 1.55 , the wavelength associated with this lowest order

destructive interference is

λ1 =

24.18

2ncoating t

= 2 ( 1.55 )( 177.4 nm ) = 550 nm

1

Since nair < noil < nwater , light reflected from both top and bottom surfaces of the oil film

experiences phase reversal, resulting in zero net phase difference due to reflections.

Therefore, the condition for constructive interference in reflected light is

2 t = mλn = m

λ

n film

λ

, or t = m

2 n film

where m = 0, 1, 2,…

Assuming that m = 1 , the thickness of the oil slick is

t = (1)

24.19

λ

2 n film

=

600 nm

= 233 nm

2 ( 1.29 )

There will be a phase reversal of the radar waves reflecting from both surfaces of the

polymer, giving zero net phase change due to reflections. The requirement for

destructive interference in the reflected waves is then

1

λ

where m = 0, 1, 2,…

2 t = m + λn , or t = ( 2m + 1)

4 n film

2

If the film is as thin as possible, then m = 0 and the needed thickness is

t=

λ

4 n film

=

3.00 cm

= 0.500 cm

4 ( 1.50 )

This anti-reflectance coating could be easily countered by changing the wavelength of

the radar—to 1.50 cm—now creating maximum reflection!

314

CHAPTER 24

24.20

The transmitted light is brightest when the reflected light is a minimum (that is, the

same conditions that produce destructive interference in the reflected light will produce

constructive interference in the transmitted light). As light enters the air layer from

glass, any light reflected at this surface has zero phase change. Light reflected from the

other surface of the air layer (where light is going from air into glass) does have a phase

reversal. Thus, the condition for destructive interference in the light reflected from the

air film is 2 t = m λn , m = 0, 1, 2,…

Since λn =

λ

n film

condition is

24.21

=

λ

1.00

= λ , the minimum non-zero plate separation satisfying this

d = t = ( 1)

λ

2

=

580 nm

= 290 nm

2

(a) For maximum transmission, we want destructive interference in the light reflected

from the front and back surfaces of the film.

If the surrounding glass has refractive index greater than 1.378, light reflected from

the front surface suffers no phase reversal, and light reflected from the back does

undergo phase reversal. This effect by itself would produce destructive interference,

so we want the distance down and back to be one whole wavelength in the film.

Thus, we require that

2t = λn = λ n film or

t=

λ

2 n film

=

656.3 nm

= 238 nm

2 ( 1.378 )

(b) The filter will expand. As t increases in 2n film t = λ , so does λ increase

(c) Destructive interference for reflected light happens also for λ in 2 t = 2 λ n film , or

λ = n filmt = ( 1.378 )( 238 nm ) = 328 nm (near ultraviolet)

Wave Optics

24.22

315

Light reflecting from the lower surface of the air layer experiences phase reversal, but

light reflecting from the upper surface of the layer does not. The requirement for a dark

fringe (destructive interference) is then

λ

2 t = mλn = m

= m λ , where m = 0, 1, 2,…

nair

At the thickest part of the film ( t = 2.00 µ m ) , the order number is

m=

2t

λ

=

2 ( 2.00 × 10 −6 m )

546.1 × 10 −9 m

= 7.32

Since m must be an integer, m = 7 is the order of the last dark fringe seen. Counting the

m = 0 order along the edge of contact, a total of 8 dark fringes will be seen.

24.23

With a phase reversal upon reflection from the

lower surface of the air layer and no phase change

for reflection at the upper surface of the layer, the

condition for destructive interference is

λ

2 t = mλn = m

= m λ , where m = 0, 1, 2,…

nair

Counting the zeroth order along the edge of

contact, the order number of the thirtieth dark

fringe observed is m = 29 . The thickness of the air

layer at this point is t = 2r , where r is the radius of

the wire. Thus,

−9

t 29 λ 29 ( 600 × 10 m )

=

= 4.35 µ m

r= =

2

4

4

Incident

light

n

t3

t2

t1

O

316

CHAPTER 24

h=Rt

24.24

R

Glass

air

no phase reversal

phase reversal

r

t{

Glass

From the geometry shown in the figure, R2 = ( R − t ) + r 2 , or

2

t = R − R2 − r 2

= 3.0 m −

2

( 3.0 m ) − ( 9.8 × 10 −3 m )

2

= 1.6 × 10 −5 m

With a phase reversal upon reflection at the lower surface of the air layer, but no

reversal with reflection from the upper surface, the condition for a bright fringe is

1

2 t = m + λn = m +

2

1 λ

= m+

2 nair

1

λ , where m = 0, 1, 2,…

2

At the 50 th bright fringe, m = 49 , and the wavelength is found to be

2 ( 1.6 × 10 −5 m )

2t

λ=

=

= 6.5 × 10 −7 m = 6.5 × 10 2 nm

m+1 2

49.5

24.25

There is a phase reversal due to reflection at the bottom of the air film but not at the top

of the film. The requirement for a dark fringe is then

2 t = mλn = m

λ

nair

= mλ , where m = 0, 1, 2,…

At the 19th dark ring (in addition to the dark center spot), the order number is m = 19 ,

and the thickness of the film is

mλ 19 ( 500 × 10

=

t=

2

2

−9

m)

= 4.75 × 10 −6 m = 4.75 µ m

Wave Optics

24.26

317

With a phase reversal due to reflection at each surface of the magnesium fluoride layer,

there is zero net phase difference caused by reflections. The condition for destructive

interference is then

2t = m +

1

λn = m +

2

1 λ

, where m = 0, 1, 2,…

2 n film

For minimum thickness, m = 0 , and the thickness is

t = ( 2 m + 1)

24.27

λ

4 n film

( 550 × 10

= ( 1)

−9

4 ( 1.38 )

m)

= 9.96 × 10 −8 m = 99.6 nm

There is a phase reversal upon reflection at each surface of the film and hence zero net

phase difference due to reflections. The requirement for constructive interference in the

reflected light is then

2 t = mλn = m

λ

n film

, where m = 1, 2, 3, …

With t = 1.00 × 10 −5 cm = 100 nm , and n film = 1.38 , the wavelengths intensified in the

reflected light are

λ=

2 n film t

m

=

2 ( 1.38 )( 100 nm )

, with m = 1, 2, 3, …

m

Thus, λ = 276 nm, 138 nm, 92.0 nm …

and none of these wavelengths are in the visible spectrum

318

CHAPTER 24

24.28

As light emerging from the glass reflects from the top of the air layer, there is no phase

reversal produced. However, the light reflecting from the end of the metal rod at the

bottom of the air layer does experience phase reversal. Thus, the condition for

constructive interference in the reflected light is 2 t = ( m + 12 ) λair .

As the metal rod expands, the thickness of the air layer decreases. The increase in the

length of the rod is given by

∆L = ∆t = ( mi +

1

2

)

λair

2

(

− mf +

1

2

) λ2

air

= ∆m

λair

2

The order number changes by one each time the film changes from bright to dark and

back to bright. Thus, during the expansion, the measured change in the length of the rod

is

∆L = ( 200 )

λair

2

= ( 200 )

( 500 × 10

2

−9

m)

= 5.00 × 10 −5 m

From ∆L = L0α ( ∆T ) , the coefficient of linear expansion of the rod is

α=

24.29

∆L

5.00 × 10 −5 m

−1

=

= 20.0 × 10 −6 ( °C )

L0 ( ∆T ) ( 0.100 m ) ( 25.0 °C )

The distance on the screen from the center to either edge of the central maximum is

λ

y = L tan θ ≈ L sin θ = L

a

632.8 × 10 −9 m

−3

= ( 1.00 m )

= 2.11 × 10 m=2.11 mm

−3

0.300 × 10 m

The full width of the central maximum on the screen is then

2 y = 4.22 mm

Wave Optics

24.30

319

(a) Dark bands occur where sin θ = m ( λ a ) . At the first dark band, m = 1 , and the

distance from the center of the central maximum is

λ

y1 = L tan θ ≈ L sin θ = L

a

600 × 10 −9 m

−3

= ( 1.5 m )

= 2.25 × 10 m = 2.3 mm

−3

0.40 × 10 m

(b) The width of the central maximum is 2 y1 = 2 ( 2.25 mm ) = 4.5 mm

24.31

(a) Dark bands (minima) occur where sin θ = m ( λ a ) . For the first minimum, m = 1 and

the distance from the center of the central maximum is y1 = L tan θ ≈ L sin θ = L ( λ a ) .

Thus, the needed distance to the screen is

0.75 × 10 −3 m

a

L = y1 = ( 0.85 × 10 −3 m )

= 1.1 m

-9

λ

587.5 × 10 m

(b) The width of the central maximum is 2 y1 = 2 ( 0.85 mm ) = 1.7 mm

Note: The small angle approximation does not

work well in this situation. Rather, you should

proceed as follows.

At the first order minimum, sin θ = λ a or

λ

5.00 cm

θ = sin −1 = sin −1

= 7.98°

a

36.0 cm

a = 36.0 cm = 0.360 m

L = 6.50 m

24.32

Central

Maximum

Then, y1 = L tan θ = ( 6.50 m ) tan 7.98° = 0.912 m = 91.2 cm

q

y1 First Order

Minimum

320

CHAPTER 24

24.33

The locations of the dark fringes (minima) mark the edges of the maxima, and the

widths of the maxima equals the spacing between successive minima.

At the locations of the minima, sin θ m = m ( λ a ) and

λ

ym = L tan θ m ≈ L sin θ m = m L

a

500 × 10 −9 m

= m ( 1.20 m )

= m ( 1.20 mm )

-3

0.500 × 10 m

Then, ∆y = ∆m ( 1.20 mm ) and for successive minima, ∆m = 1 .

Therefore, the width of each maxima, other than the central maximum, in this interference

pattern is

width = ∆y = ( 1)( 1.20 mm ) = 1.20 mm

24.34

At the positions of the minima, sin θ m = m ( λ a ) and

ym = L tan θ m ≈ L sin θ m = m L ( λ a )

Thus, y3 − y1 = ( 3 − 1) L ( λ a ) = 2 L ( λ a )

and

24.35

2 ( 0.500 m ) ( 680 × 10 −9 m )

2Lλ

a=

=

= 2.27 × 10 −4 m = 0.227 mm

−3

y3 − y1

3.00 × 10 m

The grating spacing is d =

1

1

cm =

m and d sin θ = mλ

3 660

3.66 × 10 5

(a) The wavelength observed in the first-order spectrum is λ = d sin θ , or

10 4 nm

1 m 109 nm

sin

θ

=

sin θ

5

3.66 × 10 1 m

3.66

λ =

This yields:

at 10.1°, λ = 479 nm ;

and

at 14.8°, λ = 698 nm

at 13.7°, λ = 647 nm ;

Wave Optics

321

(b) In the second order, m = 2 . The second order images for the above wavelengths will

be found at angles θ 2 = sin −1 ( 2 λ d ) = sin −1 [ 2sin θ1 ]

24.36

This yields:

for λ = 479 nm , θ 2 = 20.5° ;

and

for λ = 698 nm , θ 2 = 30.7°

for λ = 647 nm , θ 2 = 28.3° ;

(a) The longest wavelength in the visible spectrum is 700 nm, and the grating spacing is

1 mm

d=

= 1.67 × 10 −3 mm = 1.67 × 10 −6 m

600

Thus, mmax =

d sin 90.0°

λred

(1.67 × 10

=

−6

m ) sin 90.0°

700 × 10 −9 m

= 2.38

so 2 complete orders will be observed.

(b) From λ = d sin θ , the angular separation of the red and violet edges in the first order

will be

−9

700 × 10 −9 m

m

λ

λ

−1 400 × 10

∆θ = sin −1 red − sin −1 violet = sin −1

−

sin

−6

−6

d

d

1.67 × 10 m

1.67 × 10 m

or ∆θ = 10.9°

24.37

1 cm

1m

=

. From d sin θ = mλ , the angular separation

4 500 4.50 × 10 5

between the given spectral lines will be

The grating spacing is d =

m λred

m λviolet

∆θ = sin −1

− sin −1

d

d

or

m ( 656 × 10 −9 m )( 4.50 × 10 5 )

m ( 434 × 10 −9 m )( 4.50 × 10 5 )

− sin −1

∆θ = sin −1

1m

1m

The results obtained are: for m = 1, ∆θ = 5.91° ; for m = 2, ∆θ = 13.2° ;

and for m = 3, ∆θ = 26.5° . Complete orders for m ≥ 4 are not visible.

322

CHAPTER 24

24.38

(a) If d =

1 cm

= 6.67 × 10 −4 cm = 6.67 × 10 −6 m , the highest order of λ = 500 nm that can

1 500

be observed will be

mmax =

(b) If d =

λ

=

( 6.67 × 10

−6

500 × 10

m ) ( 1)

−9

m

= 13.3 or 13 orders

1 cm

= 6.67 × 10 −5 cm = 6.67 × 10 −7 m , then

15 000

mmax =

24.39

d sin 90°

d sin 90°

λ

( 6.67 × 10

=

−7

m ) ( 1)

500 × 10 −9 m

= 1.33 or 1 order

1 cm

= 2.00 × 10 −4 cm = 2.00 × 10 −6 m , and d sin θ = mλ gives

5 000

the angular position of a second order spectral line as

The grating spacing is d =

sin θ =

2λ

2λ

or θ = sin −1

d

d

For the given wavelengths, the angular positions are

2 ( 610 × 10 −9 m )

2 ( 480 × 10 −9 m )

−1

= 28.7°

θ1 = sin

= 37.6° and θ 2 = sin

−6

−6

2.00 × 10 m

2.00 × 10 m

−1

If L is the distance from the grating to the screen, the distance on the screen from the

central maximum to a second order bright line is y = L tan θ . Therefore, for the two

given wavelengths, the screen separation is

∆y = L [ tan θ1 − tan θ 2 ]

= ( 2.00 m ) tan ( 37.6° ) − tan ( 28.7° ) = 0.445 m = 44.5 cm

323

Wave Optics

24.40

With 2 000 lines per centimeter, the grating spacing is

d=

1

cm = 5.00 × 10 −4 cm = 5.00 × 10 −6 m

2 000

Then, from d sin θ = mλ , the location of the first order for the red light is

( 1) ( 640 × 10 −9 m )

mλ

−1

= 7.35°

θ = sin

= sin

−6

d

5.00 × 10 m

−1

24.41

The grating spacing is d =

1 cm 10 −2 m

=

= 3.636 × 10 −6 m . From d sin θ = mλ , or

2 750

2 750

θ = sin −1 ( mλ d ) , the angular positions of the red and violet edges of the second-order

spectrum are found to be

2 ( 700 × 10 −9 m )

2λred

−1

= 22.65°

θ r = sin

= sin

3.636 × 10 −6 m

d

−1

and

2λviolet

d

θ v = sin −1

2 ( 400 × 10 −9 m )

−1

= 12.71°

sin

=

3.636 × 10 −6 m

Note from the sketch at the right that yr = L tan θ r

and yv = L tan θ v , so the width of the spectrum on

Grating

Screen

Dy

the screen is ∆y = L ( tan θ r − tan θ v ) .

Since it is given that ∆y = 1.75 cm , the distance

from the grating to the screen must be

L=

or

∆y

1.75 cm

=

tan θ r − tan θ v tan ( 22.65° ) − tan ( 12.71° )

L = 9.13 cm

qv

L

qr

yv

yr

324

CHAPTER 24

24.42

The grating spacing is d =

1 cm

= 8.33 × 10 −4 cm = 8.33 × 10 −6 m

1 200

mλ

and the small angle approximation, the distance from the central

d

maximum to the maximum of order m for wavelength λ is

ym = L tan θ ≈ L sin θ = ( λ L d ) m . Therefore, the spacing between successive maxima is

Using sin θ =

∆y = ym+1 − ym = λ L d .

The longer wavelength in the light is found to be

λlong =

( ∆y ) d = ( 8.44 × 10−3 m )( 8.33 × 10−6 m ) =

0.150 m

L

469 nm

Since the third order maximum of the shorter wavelength falls halfway between the

central maximum and the first order maximum of the longer wavelength, we have

3 λshort L 0 + 1 λlong L

1

or λshort = ( 469 nm ) = 78.1 nm

=

d

2 d

6

24.43

The grating spacing is d =

1 mm

= 2.50 × 10 −3 mm = 2.50 × 10 −6 m

400

From d sin θ = mλ , the angle of the second-order diffracted ray is θ = sin −1 ( 2λ d ) .

(a) When the grating is surrounded by air, the wavelength is λair = λ nair ≈ λ and

2 ( 541 × 10 −9 m )

2 λair

−1

= 25.6°

θ a = sin

= sin

−6

d

2.50 × 10 m

−1

(b) If the grating is immersed in water,

then λn = λwater =

λ

nwater

2 λwater

d

θ b = sin −1

=

λ

1.333

, yielding

2 ( 541 × 10 −9 m )

−1

= 19.0°

sin

=

−6

( 2.50 × 10 m ) ( 1.333 )

mλn m ( λ n )

mλ

=

, we have that n sin θ =

= constant when m is kept

d

d

d

constant. Therefore nair sin θ a = nwater sin θ b , or the angles of parts (a) and (b) satisfy

(c) From sin θ =

Snell’s law.

Wave Optics

24.44

When light of wavelength λ passes through a single slit of width a, the first minimum is

observed at angle θ where

sin θ =

λ

λ

or θ = sin −1

a

a

This will have no solution if a < λ , so the maximum slit width if no minima are to be

seen is a = λ = 632 .8 nm .

24.45

(a) From Brewster’s law, the index of refraction is

n2 = tan θ p = tan ( 48.0° ) = 1.11

(b) From Snell’s law, n2 sin θ 2 = n1 sin θ1 , we obtain when θ1 = θ p

n1 sin θ p

−1 ( 1.00 ) sin 48.0°

= sin

= 42.0°

1.11

n2

θ 2 = sin −1

Note that when θ1 = θ p , θ 2 = 90.0° − θ p as it should.

24.46

Unpolarized light incident on a polarizer contains electric field vectors at all angles to

the transmission axis of the polarizer. Malus’s law then gives the intensity of the

transmitted light as I = I 0 ( cos 2 θ )av . Since the average value of cos 2 θ is 1 2 , the

intensity of the light passed by the first polarizer is I1 = I 0 2 , where I 0 is the incident

intensity.

Then, from Malus’s law, the intensity passed by the second polarizer is

I

3

I 3

I 2 = I1 cos 2 ( 30.0° ) = 0 , or 2 =

I0

8

2 4

24.47

325

The more general expression for Brewster’s angle is (see Problem 51)

tan θ p = n2 n1

n

1.52

(a) When n1 = 1.00 and n2 = 1.52 , θ p = tan −1 2 = tan −1

= 56.7°

1.00

n1

(b)

n

When n1 = 1.333 and n2 = 1.52 , θ p = tan −1 2

n1

−1 1.52

= tan

= 48.8°

1.333

326

CHAPTER 24

24.48

The polarizing angle for light in air striking a water surface is

n2

n1

θ p = tan −1

−1 1.333

= tan

= 53.1°

1.00

This is the angle of incidence for the incoming sunlight (that is, the angle between the

incident light and the normal to the surface). The altitude of the Sun is the angle

between the incident light and the water surface. Thus, the altitude of the Sun is

α = 90.0° − θ p = 90.0° − 53.1° = 36.9°

24.49

n

1.65

The polarizing angle is θ p = tan −1 2 = tan −1

= 58.8°

1.00

n1

When the light is incident at the polarizing angle, the angle of refraction is

θ r = 90.0° − θ p = 90.0° − 58.8° = 31.2°

24.50

The critical angle for total reflection is θ c = sin −1 ( n2 n1 ) . Thus, if θ c = 34.4° as light

attempts to go from sapphire into air, the index of refraction of sapphire is

nsapphire = n1 =

n2

1.00

=

= 1.77

sin θ c sin 34.4°

Then, when light is incident on sapphire from air, the Brewster angle is

n2

−1 1.77

= tan

= 60.5°

1.00

n1

θ p = tan −1

24.51

From Snell’s law, the angles of incidence and refraction are related by n1 sin θ1 = n2 sin θ 2 .

If the angle of incidence is the polarizing angle (that is, θ1 = θ p ), the angles of incidence

and refraction are also related by

θ p + θ 2 + 90° = 180° , or θ 2 = 90° − θ p

Substitution into Snell’s law then gives

(

)

n1 sin θ p = n2 sin 90° − θ p = n2 cosθ p or tan θ p = n2 n1

Wave Optics

24.52

24.53

I = I 0 cos 2 θ

I

I0

⇒

θ = cos −1

(a)

I

1

=

⇒

I 0 3.00

θ = cos −1

(b)

I

1

=

⇒

I 0 5.00

θ = cos −1

(c)

I

1

=

I 0 10.0

θ = cos −1

⇒

1

= 54.7°

3.00

1

= 63.4°

5.00

1

= 71.6°

10.0

From Malus’s law, the intensity of the light transmitted by the first polarizer is

I1 = I i cos 2 θ1 . The plane of polarization of this light is parallel to the axis of the first plate

and is incident on the second plate. Malus’s law gives the intensity transmitted by the

second plate as I 2 = I1 cos 2 (θ 2 − θ1 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) . This light is polarized parallel

to the axis of the second plate and is incident upon the third plate. A final application of

Malus’s law gives the transmitted intensity as

I f = I 2 cos 2 (θ 3 − θ 2 ) = I i cos 2 θ1 cos 2 (θ 2 − θ1 ) cos 2 (θ 3 − θ 2 )

With θ1 = 20.0°, θ 2 = 40.0°, and θ 3 = 60.0° , this result yields

I f = ( 10.0 units ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) cos 2 ( 20.0° ) = 6.89 units

24.54

327

(a) Using Malus’s law, the intensity of the transmitted light is found to be

(

I = I 0 cos 2 ( 45° ) = I 0 1

)

2

2 , or I I 0 = 1 2

(b) From Malus’s law, I I 0 = cos 2 θ . Thus, if I I 0 = 1 3 we obtain

(

cos 2 θ = 1 3 or θ = cos −1 1

)

3 = 54.7°

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