sm chapter38

38Diffraction Patterns and Polarization

CHAPTER OUTLINE

38.1 Introduction to Diffraction Patterns

38.2 Diffraction Patterns from Narrow Slits

38.3 Resolution of Single-Slit and Circular Apertures

38.4 The Diffraction Grating38.5 Diffraction of X-Rays by Crystals38.6 Polarization of Light Waves

ANSWERS TO QUESTIONS

Q38.1 Audible sound has wavelengths on the order of meters or centimeters, while visible light has a wavelength on the order of half a micrometer. In this world of

breadbox-sized objects, λa

is large for sound, and sound

diffracts around behind walls with doorways. But λa

is a tiny fraction for visible light passing ordinary-size objects or apertures, so light changes its direction by only very small angles when it diffracts.

Another way of phrasing the answer: We can see by a small angle around a small obstacle or around the edge of a small opening. The side fringes in Figure 38.1 and the Arago spot in the center of Figure 38.3 show this diffraction. We cannot always hear around corners. Out-of-doors, away from refl ecting surfaces, have some-one a few meters distant face away from you and whisper. The high-frequency, short-wavelength, information-carrying components of the sound do not diffract around his head enough for you to understand his words.

Suppose an opera singer loses the tempo and cannot immediately get it from the orchestra conductor. Then the prompter may make rhythmic kissing noises with her lips and teeth. Try it—you will sound like a birdwatcher trying to lure out a curious bird. This sound is clear on the stage but does not diffract around the prompter’s box enough for the audience to hear it.

Q38.2 The wavelength of visible light is extremely small in comparison to the dimensions of your hand, so the diffraction of light around an obstacle 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, signifi cant diffraction of sound waves occurs around hand-sized obstacles.

*Q38.3 Answer (d). The power of the light coming through the slit decreases, as you would expect. The

central maximum increases in width as the width of the slit decreases. In the condition sinθ λ=a

for destructive interference on each side of the central maximum, θ increases as a decreases.

*Q38.4 We consider the fraction λL/a. In case (a) and in case (f) it is λ0L/a. In case (b) it is 2λ

0L/3a.

In case (c) it is 3λ0L/2a. In case (d) it is l

0L/2a. In case (e) it is l

02L/a. The ranking is e > c > a =

f > b > d.

*Q38.5 Answer (b). The wavelength will be much smaller than with visible light.

379

ISMV2_5104_ch38.indd 379ISMV2_5104_ch38.indd 379 6/30/07 8:35:44 PM6/30/07 8:35:44 PM

380 Chapter 38

*Q38.6 Answer (c). The ability to resolve light sources depends on diffraction, not on intensity.

Q38.7 Consider incident light nearly parallel to the horizontal ruler. Suppose it scatters from bumps at distance d apart to produce a diffraction pattern on a vertical wall a

distance L away. At a point of height y, where θ = y

L gives the scattering angle θ, the character of the

interference is determined by the shift δ between beams

scattered by adjacent bumps, where δθ

= −dd

cos.

Bright spots appear for δ λ= m , where m = 0, 1, 2, 3, . . . . For small θ, these equations combine

and reduce to my d

Lmλ =2

22. Measurement of the heights ym of bright spots allows calculation of the

wavelength of the light.

*Q38.8 Answer (b). No diffraction effects are observed because the separation distance between adjacent ribs is so much greater than the wavelength of x-rays. Diffraction does not limit the resolution of an x-ray image. Diffraction might sometimes limit the resolution of an ultrasonogram.

*Q38.9 Answer (a). Glare, as usually encountered when driving or boating, is horizontally polarized. Refl ected light is polarized in the same plane as the refl ecting surface. As unpolarized light hits a shiny horizontal surface, the atoms on the surface absorb and then reemit the light energy as a refl ection. We can model the surface as containing conduction electrons free to vibrate eas-ily along the surface, but not to move easily out of surface. The light emitted from a vibrating electron is partially or completely polarized along the plane of vibration, thus horizontally.

Q38.10 Light from the sky is partially polarized. Light from the blue sky that is polarized at 90° to the polarization axis of the glasses will be blocked, making the sky look darker as compared to the clouds.

*Q38.11 Answer (a). The grooves in a diffraction grating are not electrically conducting. Sending light through a diffraction grating is not like sending a vibration on a rope through a picket fence. The electric fi eld in light does not have an amplitude in real space. Its amplitude is in newtons per coulomb, not in millimeters.

Q38.12 First think about the glass without a coin and about one particular point P on the screen. We can divide up the area of the glass into ring-shaped zones centered on the line joining P and the light source, with successive zones contributing alternately in-phase and out-of-phase with the light that takes the straight-line path to P. These Fresnel zones have nearly equal areas. An outer zone contributes only slightly less to the total wave disturbance at P than does the central circular zone. Now insert the coin. If P is in line with its center, the coin will block off the light from some particular number of zones. The fi rst unblocked zone around its circumference will send light to P with signifi cant amplitude. Zones farther out will predominantly interfere destructively with each other, and the Arago spot is bright. Slightly off the axis there is nearly complete destructive interference, so most of the geometrical shadow is dark. A bug on the screen crawling out past the edge of the geometrical shadow would in effect see the central few zones coming out of eclipse. As the light from them interferes alternately constructively and destructively, the bug moves through bright and dark fringes on the screen. The diffraction pattern is shown in Figure 38.3 in the text.

FIG. Q38.7

q

d L

y


Diffraction Patterns and Polarization 381

Q38.13 Since the obsidian is opaque, a standard method of measuring incidence and refraction angles and using Snell’s Law is ineffective. Refl ect unpolarized light from the horizontal surface of the obsidian through a vertically polarized fi lter. Change the angle of incidence until you observe that none of the refl ected light is transmitted through the fi lter. This means that the refl ected light is completely horizontally polarized, and that the incidence and refl ection angles are the polarization angle. The tangent of the polarization angle is the index of refraction of the obsidian.

Q38.14 The fi ne hair blocks off light that would otherwise go through a fi ne slit and produce a diffraction pattern on a distant screen. The width of the central maximum in the pattern is inversely proportional to the distance across the slit. When the hair is in place, it subtracts the same diffraction pattern from the projected disk of laser light. The hair produces a diffraction minimum that crosses the bright circle on the screen. The width of the minimum is inversely proportional to the diameter of the hair. The central minimum is fl anked by narrower maxima and minima. Measure the width 2y of the central minimum between the maxima bracketing it, and use

Equation 38.1 in the form y

L a= λ to fi nd the width a of the hair.

Q38.15 The condition for constructive interference is that the three radio signals arrive at the city in phase. We know the speed of the waves (it is the speed of light c), the angular bearing θ of the city east of north from the broadcast site, and the distance d between adjacent towers. The wave from the westernmost tower must travel an extra distance 2d sinθ to reach the city, compared to the signal from the eastern tower. For each cycle of the carrier wave, the western antenna would

transmit fi rst, the center antenna after a time delay d

c

sinθ , and the eastern antenna after an additional equal time delay.

Q38.16 It is shown in the correct orientation. If the horizontal width of the opening is equal to or less than the wavelength of the sound, then the equation a sinθ λ= ( )1 has the solution θ = °90 , or has no solution. The central diffraction maximum covers the whole seaward side. If the vertical height of the opening is large compared to the wavelength, then the angle in a sinθ λ= ( )1 will be small, and the central diffraction maximum will form a thin horizontal sheet.

Featured in the motion picture M*A*S*H (20th Century Fox, Aspen Productions, 1970) is a loudspeaker mounted on an exterior wall of an Army barracks. It has an approximately rectangular aperture, and it is installed incorrectly. The longer side is horizontal, to maximize sound spreading in a vertical plane and to minimize sound radiated in different horizontal directions.

SOLUTIONS TO PROBLEMS

Section 38.1 Introduction to Diffraction Patterns

Section 38.2 Diffraction Patterns from Narrow Slits

P38.1 sin.

..θ λ= = ×

×= ×

−

−−

a

6 328 10

3 00 102 11 10

7

43

y

1 00.tan sin

m= ≈ ≈θ θ θ (for small θ)

2 4 22y = . mm


382 Chapter 38

P38.2 The positions of the fi rst-order minima are y

L a≈ = ±sinθ λ

. Thus, the spacing between these two

minima is ∆ya

L= ⎛⎝⎜

⎞⎠⎟2

λ and the wavelength is

λ = ⎛⎝

⎞⎠

⎛⎝

⎞⎠ = ×⎛

⎝⎜⎞⎠⎟

×− −∆y a

L2

4 10 10 0 550 103. .m

2

33

547m

2.06 mnm

⎛⎝⎜

⎞⎠⎟

=

P38.3 y

L

m

a≈ =sinθ λ

∆y = × −3 00 10 3. nm

∆m = − =3 1 2 and am L

y= ∆

∆λ

a =×( )( )

×( ) = ×−

−

2 690 10 0 500

3 00 102 30

9

3

m m

m

.

.. 110 4− m

P38.4 For destructive interference,

sin.

.θ λ λ= = = =ma a

5 000 139

cm

36.0 cm

and θ = °7 98.

y

L= tanθ

gives y L= = ( ) =tan . tan . .θ 6 50 7 98 0 912m m°

y = 91 2. cm

P38.5 If the speed of sound is 340 m/s, λ = = =−

vf

340

6500 5231

m s

sm.

Diffraction minima occur at angles described by a msinθ λ=

1 10 1 0 5231. sin .m m( ) = ( )θ θ1 28 4= °.

1 10 2 0 5232. sin .m m( ) = ( )θ θ2 72 0= °.

1 10 3 0 5233. sin .m m( ) = ( )θ θ3 nonexistent

Maxima appear straight ahead at 0° and left and right at an angle given approximately by

1 10 1 5 0 523. sin . .m m( ) = ( )θx θx ≈ °46

There is no solution to a sin .θ λ= 2 5 , so our answer is already complete, with three sound maxima.



*P38.6 (a) The rectangular patch on the wall is wider than it is tall. The aperture will be taller than it is wide. For horizontal spreading we have

tan.

..θwidth

width

widt

m 2

m= = =

y

La

0 110

4 50 012 2

hh width

width

m

0.012 2

sin

..

θ λ=

= × =−

1

632 8 105 1

9

a 88 10 5× − m

For vertical spreading, similarly

tan.

..θ

λ

height

height

m 2

m= =

=

0 006

4 50 000 667

1a

ssin

.

.

θh

= ×

= ×

−

−

632 8 10

9 49 10

9

4

m

0.000 667

m

(b) The central bright patch is horizontal. The aperture is vertical.

A smaller distance between aperture edges causes a wider diffraction angle. The longer dimension of each rectangle is 18.3 times larger than the smaller dimension.

P38.7 sin.θ ≈ = × −y

L

4 10 10 3 m

1.20 m

We definem

φ π θλ

π= =

×( )×

−

−

a sin .

.

4 00 10

546 1 10

4

99

34 10 107 86

m

m

1.20 mrad

..

s

max

×⎛⎝⎜

⎞⎠⎟

=

=

−

I

I

iin sin .

..

φφ( )⎡

⎣⎢

⎤

⎦⎥ = ( )⎡

⎣⎢⎤⎦⎥

= ×2 2

7 86

7 861 62 10−−2

P38.8 Equation 38.1 states that sinθ λ= m

a, where m = ± ± ±1 2 3, , , . . . .

The requirement for m = 1 is from an analysis of the extra path distance traveled by ray 1 compared to ray 3 in the textbook

Figure 38.5. This extra distance must be equal to λ2

for destructive

interference. When the source rays approach the slit at an angle b, there is a distance added to the path difference (of ray 1 compared to ray 3) of

a

2sin β. Then, for destructive interference,

a a

2 2 2sin sinβ θ λ+ =

so sin sinθ λ β= −

a0 .

Dividing the slit into 4 parts leads to the 2nd order minimum: sin sinθ λ β= −2

a

Dividing the slit into 6 parts gives the third order minimum: sin sinθ λ β= −3

a

Generalizing, we obtain the condition for the mth order minimum: sin sinθ λ β= −m

a

FIG. P38.6

FIG. P38.8


384 Chapter 38

*P38.9 The diffraction envelope shows a broad central maximum fl anked by zeros at a sin θ = 1l and a sin θ = 2l. That is, the zeros are at (π a sin θ)/λ = π , −π , 2π , −2π , . . . . Not-ing that the distance between slits is d = 9µm = 3a, we say that within the diffraction envelope the interference pattern shows closely spaced maxima at d sin θ = mλ, giving (π 3a sin θ)/λ = mπ or (πa sin θ)/λ = 0, π /3, −π /3, 2π /3, −2π /3. The third-order interference maxima are missing because they fall at the same directions as diffraction minima, but the fourth order can be visible at (πa sin θ)/λ = 4π /3 and –4π /3 as diagrammed.

P38.10 (a) Double-slit interference maxima are at angles given by d msinθ λ= .

For m = 0, θ0 0= °

For m = 1, 2 80 1 0 501 5. sin .m mµ θ µ( ) = ( ): θ11 0 179 10 3= ( ) = °−sin . .

Similarly, for m = 2 3 4 5, , , and 6, θ2 21 0= °. , θ3 32 5= °.

, θ4 45 8= °.

θ5 63 6= °. , and θ6

1 1 07= ( ) =−sin . nonexistent

Thus, there are 5 5 1 11+ + = directions for interference maxima.

(b) We check for missing orders by looking for single-slit diffraction minima, at a msinθ λ= .

For m = 1, 0 700 1 0 501 5. sin .m mµ θ µ( ) = ( ) and θ1 45 8= °.

Thus, there is no bright fringe at this angle. There are only nine bright fringes , at

θ = ° ± ° ± ° ± ° ± °0 10 3 21 0 32 5 63 6, . , . , . , .and .

(c) I Ia

= ( )⎡⎣⎢

⎤⎦⎥max

sin sin

sin

π θ λπ θ λ

2

At θ = °0 , sinθ

θ→ 1 and

I

Imax

.→ 1 00

At θ = °10 3. , π θ

λπ µ

µa sin . sin .

..= ( ) °

=0 700 10 3

0 501 50 785

m

mrrad = °45 0.

I

Imax

sin .

..= °⎡

⎣⎢⎤⎦⎥

=45 0

0 7850 811

2

Similarly, at θ = °21 0. , π θ

λa sin

. .= = °1 57 90 0rad and

I

Imax

.= 0 405

At θ = °32 5. , π θ

λa sin

.= = °2 36 135rad and I

Imax

.= 0 090 1

At θ = °63 6. , π θ

λa sin

.= = °3 93 225rad and I

Imax

.= 0 032 4

−p p pasinql

0

I

FIG. P38.9



Section 38.3 Resolution of Single-Slit and Circular Apertures

P38.11 We assume Rayleigh’s criterion applies to the predator’s eye with pupil narrowed. (It is a few times too optimistic for a normal human eye with pupil dilated.)

sin.

.θ λ= = ××

= ×−

−−

a

5 00 101 00 10

7

43m

5.00 10rad

*P38.12 (a) 1 22 1 22 589 9 000 000 79 8. / . ( /( .λ µD = nm) nm) = rad

(b) For a smaller angle of diffraction we choose the smallest visible wavelength, violet at

400 nm, to obtain 1 22 1 22 400 9 000 000 54 2. / . ( /( .λ µD = nm) nm) = rad .

(c) The wavelength in water is shortened to its vacuum value divided by the index of refraction.

The resolving power is improved, with the miinimum resolvable angle becoming

1 22 1 22 589 1 33 9 000 000 60 0. / . ( / . )/( .λ µD = nm nm) = raad

Better than water for many purposes is oil immersion.

P38.13 Undergoing diffraction from a circular opening, the beam spreads into a cone of half-angle

θ λmin . .

.

.= = ×⎛

⎝⎜⎞⎠⎟

=−

1 22 1 22632 8 10

0 005 001

9

D

m

m..54 10 4× − rad

The radius of the beam ten kilometers away is, from the defi nition of radian measure,

rbeam m m= ×( ) =θmin . .1 00 10 1 5444

and its diameter is d rbeam beam m= =2 3 09. .

*P38.14 When the pupil is open wide, it appears that the resolving power of human vision is limited by the coarseness of light sensors on the retina. But we use Rayleigh’s criterion as a handy indicator

of how good our vision might be. We take θ λmin .= =d

L D1 22 , where θmin is the smallest angular

separation of two objects for which they are resolved by an aperture of diameter D, d is the separation of the two objects, and L is the maximum distance of the aperture from the two objects at which they can be resolved.

Two objects can be resolved if their angular separation is greater than θmin. Thus, θmin should be as small as possible. Therefore, light with the smaller of the two given wavelengths is easier to resolve, i.e., blue .

LDd= =

×( ) ×( )=

− −

1 22

5 20 10 2 80 10

1 22

13 2

.

. .

.λ λm m ..193 10 4× − m2

λ

Thus L = 186 m for λ = 640 nm, and L = 271 m for λ = 440 nm. The viewer with the assumed

diffraction-limited vision could resolve adjacent tubes of blue in the range 186 m to 271 m , but cannot resolve adjacent tubes of red in this range.

P38.15 When the pupil is open wide, it appears that the resolving power of human vision is limited by the coarseness of light sensors on the retina. But we use Rayleigh’s criterion as a handy indicator of how good our vision might be. According to this criterion, two dots separated center-to-center by 2.00 mm would overlap

when θ λmin .= =d

L D1 22

Thus, LdD= =

×( ) ×( )− −

1 22

2 00 10 4 00 10

1 22 500

3 3

.

. .

.λm m

××( ) =−1013 1

9 mm.


386 Chapter 38

P38.16 When the pupil is open wide, it appears that the resolving power of human vision is limited by the coarseness of light sensors on the retina. But we use Rayleigh’s criterion as a handy indicator of how good our vision might be.

xd

D= = ××

⎛⎝⎜

⎞⎠⎟

−

−1 22 1 225 00 10

27

3. ..λ m

5.00 10 m550 10 30 53×( ) =m m.

P38.17 The concave mirror of the spy satellite is probably about 2 m in diameter, and is surely not more than 5 m in diameter. That is the size of the largest piece of glass successfully cast to a precise shape, for the mirror of the Hale telescope on Mount Palomar. If the spy satellite had a larger mir-ror, its manufacture could not be kept secret, and it would be visible from the ground. Outer space is probably closer than your state capitol, but the satellite is surely above 200-km altitude, for reasonably low air friction. We fi nd the distance between barely resolvable objects at a distance of 200 km, seen in yellow light through a 5-m aperture:

y

L D

y

=

= ( )( ) ×⎛⎝⎜

⎞⎠

−

1 22

200 000 1 226 10 7

.

.

λ

mm

5 m ⎟⎟ = 3 cm

(Considering atmospheric seeing caused by variations in air density and temperature, the distance

between barely resolvable objects is more like 200 000 11

3 600m s

s

rad

180( )( ) °⎛

⎝⎜⎞⎠⎟ °

⎛⎝⎜

⎞⎠

π⎟⎟ = 97 cm.)

Thus the snooping spy satellite cannot see the difference between III and II or IV on a license plate. It cannot count coins spilled on a sidewalk, much less read the dates on them.

P38.18 1 22.λD

d

L= λ = =c

f0 020 0. m

D = 2 10. m L = 9 000 m

d =( )( ) =1 220 020 0 9 000

2 10105.

.

.

m m

mm

Section 38.4 The Diffraction Grating

P38.19 d = = × =−1 00 1 00 10

5 002. .

.cm

2 000

m

2 000mµ

sin.

.θ λ= =×( )×

=−

−

m

d

1 640 10

5 00 100 128

9

6

m

m θ = °7 35.

D

d

= ×

= ×

= ×

−

−

250 10

5 00 10

5 00 10

3

7

3

m

m

m

λ .

.



P38.20 The principal maxima are defi ned by

d msinθ λ= m = 0 1 2, , , . . .

For m = 1, λ θ= d sin

where q is the angle between the central (m = 0) and the fi rst order (m = 1) maxima. The value of q can be determined from the information given about the distance between maxima and the grating-to-screen distance. From the fi gure,

tan.

.θ = =0 4880 284

m

1.72 m

so θ = °15 8.

and sin .θ = 0 273

The distance between grating “slits” equals the reciprocal of the number of grating lines per centimeter

d = = × = ×−−1

5 3101 88 10 1 88 101

4 3

cmcm nm. .

The wavelength is λ θ= = ×( )( ) =d sin . .1 88 10 0 273 5143 nm nm

P38.21 The grating spacing is d = × = ×−

−1 00 102 22 10

26.

.m

4 500m

In the 1st-order spectrum, diffraction angles are given by

sinθ λ=d

: sin .θ1

9

6

656 100 295= ×

×=

−

−

m

2.22 10 m

so that for red θ1 17 17= °.

and for blue sin .θ2

9

6

434 10

100 195= ×

×=

−

−

m

2.22 m

so that θ2 11 26= °.

The angular separation is in fi rst-order, ∆θ = ° − ° = °17 17 11 26 5 91. . .

In the second-order spectrum, ∆θ λ λ= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟ = °− −sin sin .1 1 1 22 2

13 2d d

Again, in the third order, ∆θ λ λ= ⎛⎝⎜

⎞⎠⎟ − ⎛

⎝⎜⎞⎠⎟ = °− −sin sin .1 1 1 23 3

26 5d d

Since the red does not appear in the fourth-order spectrum, the answer is complete.

FIG. P38.20

1.72 m

0.488 mq

FIG. P38.21


388 Chapter 38

P38.22 sin .θ = 0 350: d = = = ×λθsin

..

632 81 81 103nm

0.350nm

Line spacing = 1 81. mµ

P38.23 (a) d = = × = × =− −1

3 6602 732 10 2 732 104 6

lines cmcm m. . 22 732 nm

λ θ= d

m

sin: At θ = °10 09. , λ = 478 7. nm

At θ = °13 71. , λ = 647 6. nm

At θ = °14 77. , λ = 696 6. nm

(b) d = λθsin 1

and 2 2λ θ= d sin so sinsin

sinθ λ λλ θ

θ21

1

2 22= = =

d

Therefore, if θ1 10 09= °. then sin sin( . )θ2 2 10 09= ° gives θ2 20 51= °. .

Similarly, for θ1 13 71= °. , θ2 28 30= °. and for θ1 14 77= °. , θ2 30 66= °. .

P38.24 sinθ λ= m

d

Therefore, taking the ends of the visible spectrum to be λv = 400 nm and λr = 750 nm, the ends of the different order spectra are defi ned by:

End of second order: sinθ λ2

2 1 500r

r

d d= = nm

Start of third order: sinθ λ3

3 1 200v

v= =d d

nm

Thus, it is seen that θ θ2 3r > v and these orders must overlap regardless of the value of the

grating spacing d.

*P38.25 The sound has wavelength λ = =×

= × −vf

343

37 2 109 22 103

3m s

sm

.. . Each diffracted beam is described

by d m msinθ λ= =, , , , . . .0 1 2

The zero-order beam is at m = 0, θ = 0. The beams in the fi rst order of interference are to the

left and right at θ λ= = ××

=− −−

−−sin sin

.

.sin1 1

3

211 9 22 10

1 3 10d

m

m00 709 45 2. .= °. For a second-order beam

we would need θ λ= = ( ) =− − −sin sin . sin .1 1 122 0 709 1 42

d. No angle, smaller or larger than 90°,

has a sine greater than 1. Then a diffracted beam does not exist for the second order or any higher

order. The whole answer is then, three beams, at 0 and at 45.2 to the right° ° and to the left .



P38.26 For a side maximum, tan.θ µ

µ= =y

L

0 4 m

6.9 m

θ = °3 32.

d msinθ λ= d =( ) ×( )

°=

−1 780 10

3 3213 5

9 mm

sin .. µ

The number of grooves per millimeter = ××

=

−

−

1 10

74 2

3 m

13.5 10 m6

.

P38.27 d = × = × =−

−1 00 10

2504 00 10 4 000

36.

.m mm

lines mmm nm d m m

dsin

sinθ λ θλ

= ⇒ =

(a) The number of times a complete order is seen is the same as the number of orders in which the long wavelength limit is visible.

md

maxmaxsin sin .

.= = ( ) °=

θλ

4 000 90 0

7005 71

nm

nm

or 5 orders is the maximum

(b) The highest order in which the violet end of the spectrum can be seen is:

md

maxmaxsin sin .

.= = ( ) °=

θλ

4 000 90 0

40010 0

nm

nm

or 10 orders in the short-wavelength region

P38.28 (a) The several narrow parallel slits make a diffraction grating. The zeroth- and fi rst-order maxima are separated according to

d sinθ λ= ( )1 sin.θ λ= = ××

−

−d

632 8 10 9

3

m

1.2 10 m

θ = ( ) =−sin . .1 0 000 527 0 000 527 rad

y L= = ( )( ) =tan . . .θ 1 40 0 000 527 0 738m mm

(b) Many equally spaced transparent lines appear on the fi lm. It is itself a diffraction grating. When the same light is sent through the fi lm, it produces interference maxima separated according to

d sinθ λ= ( )1 sin.

.θ λ= = ××

=−

−d

632 8 100 000 857

9

3

m

0.738 10 m

y L= = ( )( ) =tan . . .θ 1 40 0 000 857 1 20m mm

An image of the original set of slits appears on the screen. If the screen is removed, light diverges from the real images with the same wave fronts reconstructed as the original slits produced. Reasoning from the mathematics of Fourier transforms, Gabor showed that light diverging from any object, not just a set of slits, could be used. In the picture, the slits or maxima on the left are separated by 1.20 mm. The slits or maxima on the right are separated by 0.738 mm. The length difference between any pair of lines is an integer number of wavelengths. Light can be sent through equally well toward the right or toward the left. Soccer players shift smoothly between offensive and defensive tactics.

FIG. P38.26

FIG. P38.28


390 Chapter 38

P38.29 d = = × =−1

4 2002 38 10 2 3806

cmm nm.

d msinθ λ= or θ λ= ⎛⎝⎜

⎞⎠⎟

−sin 1 m

d and y L L

m

d= = ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

−tan tan sinθ λ1

Thus, ∆y Lm

d

m

d= ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

− ⎛⎝

− −tan sin tan sin1 2 1 1λ λ⎜⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

⎧⎨⎩

⎫⎬⎭

For m = 1, ∆y = ( ) ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥ −−2 00

589 6

2 3801. tan sin

.tam nn sin .− ⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=1 589

2 3800 554 mmm

For m = 2, ∆y = ( ) ( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−2 002 589 6

2 3801. tan sin

.m −− ( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=−tan sin 1 2 589

2 38011 54. mm

For m = 3, ∆y = ( ) ( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

−2 003 589 6

2 3801. tan sin

.m −− ( )⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪=−tan sin 1 3 589

2 38055 04. mm

Thus, the observed order must be m = 2 .

Section 38.5 Diffraction of X-Rays by Crystals

P38.30 2d msinθ λ= : λ θ= =×( ) °

= ×−

−2 2 0 353 10 7 60

19 34 10

91d

m

sin . sin ..

m 11 0 093 4m nm= .

P38.31 2d msinθ λ= : sin.

..θ λ= =

×( )×( ) =

−

−

m

d2

1 0 140 10

2 0 281 100

9

9

m

m2249

and θ = °14 4.

P38.32 Figure 38.23 of the text shows the situation.

2d msinθ λ= or λ θ= 2d

m

sin

m = 1: λ1

2 2 80 80 0

15 51= ( ) ° =. sin .

.m

m

m = 2: λ2

2 2 80 80 0

22 76= ( ) ° =. sin .

.m

m

m = 3: λ3

2 2 80 80 0

31 84= ( ) ° =. sin ..

mm

*P38.33 The crystal cannot produce diffracted beams of visible light. The wavelengths of visible light are some hundreds of nanometers. There is no angle whose sine is greater than 1. Bragg’s law 2d sin q = ml cannot be satisfi ed for a wavelength much larger than the distance between atomic planes in the crystal.



Section 38.6 Polarization of Light Waves

P38.34 The average value of the cosine-squared function is one-half, so the fi rst polarizer transmits

1

2 the light. The second transmits cos .2 30 0

3

4° = .

I I If i i= × =1

2

3

4

3

8

P38.35 I I= max cos2 θ ⇒ θ = −cosmax

1 I

I

(a) I

Imax .= 1

3 00 ⇒ θ = = °−cos

..1 1

3 0054 7

(b) I

Imax .= 1

5 00 ⇒ θ = = °−cos

..1 1

5 0063 4

(c) I

Imax .= 1

10 0 ⇒ θ = = °−cos

..1 1

10 071 6

P38.36 By Brewster’s law, n p= = °( ) =tan tan . .θ 48 0 1 11

P38.37 sinθc n= 1

or nc

= =°

=1 1

34 41 77

sin sin ..

θ Also, tanθ p n= . Thus, θ p n= ( ) = ( ) = °− −tan tan . .1 1 1 77 60 5

P38.38 sinθc n= 1

and tanθ p n=

Thus, sintan

θθc

p

= 1 or cot sinθ θp c= .

P38.39 (a) At incidence, n n1 1 2 2sin sinθ θ= and ′ =θ θ1 1. For complete polarization of the refl ected light,

90 90 90

90

1 2

1 2 1 2

− ′( ) + −( ) = °

′ + = = +

θ θθ θ θ θ

Then n n n1 1 2 1 2 190sin sin cosθ θ θ= −( ) =

sin

costan

θθ

θ1

1

2

11= =n

n

At the bottom surface, θ θ3 2= because the normals to the surfaces of entry and exit are parallel.

Then n n2 3 1 4sin sinθ θ= and ′ =θ θ3 3

n n2 2 1 4sin sinθ θ= and θ θ4 1=

The condition for complete polarization of the refl ected light is

90 90 903 4− ′ + − = °θ θ θ θ2 1 90+ = °

This is the same as the condition for θ1 to be Brewster’s angle at the top surface.

continued on next page

FIG. P38.39(a)

n1

n2

q1 q1'

n1

q3q3

'

q2

q4


392 Chapter 38

(b) We consider light moving in a plane perpendicular to the line where the surfaces of the prism meet at the unknown angle Φ. We require

n n1 1 2 2

1 2 90

sin sinθ θθ θ

=+ = °

So n n1 2 2 290sin sin−( ) =θ θ n

n1

22= tanθ

And n n2 3 3 4sin sinθ θ= θ θ3 4 90+ = °

n n2 3 3 3sin cosθ θ= tanθ33

2

=n

n

In the triangle made by the faces of the prism and the ray in the prism,

Φ + + + −( ) =90 90 1802 3θ θ

So one particular apex angle is required, annd it is

Φ = − =⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

− −θ θ3 21 3

2

1 1

2

tan tann

n

n

n

Here a negative result is to be interpreted as meaning the same as a positive result.

P38.40 For incident unpolarized light of intensity Imax:

The average value of the cosine-squared function is one half, so the intensity after transmission by the fi rst disk is

I I= 1

2 max

After transmitting 2nd disk: I I= 1

22

max cos θ

After transmitting 3rd disk: I I= ° −( )1

2902 2

max cos cosθ θ

where the angle between the fi rst and second disk is θ ω= t .

Using trigonometric identities cos cos2 1

21 2θ θ= +( )

and cos sin cos2 2901

21 2° −( ) = = −( )θ θ θ

we have I I= +( )⎡⎣⎢

⎤⎦⎥

−( )⎡⎣⎢

⎤⎦⎥

1

2

1 2

2

1 2

2max

cos cosθ θ

I I I= −( ) = ⎛⎝

⎞⎠ −( )1

81 2

1

8

1

21 42

max maxcos cosθ θ

Since θ ω= t, the intensity of the emerging beam is given by

I I t= −( )1

161 4max ω .

FIG. P38.39(b)

q1 q1

q2 q3 q3

q4

Φn1

n2

n3

FIG. P38.40



*P38.41 (a) Let I0 represent the intensity of unpolarized light incident on the fi rst polarizer. In Malus’s

law the average value of the cosine-squared function is 1/2, so the fi rst fi lter lets through 1/2 of the incident intensity. Of the light reaching them, the second fi lter passes cos245° = 1/2 and the third fi lter also cos245° = 1/2. The transmitted intensity is then I

0(1/2)(1/2)(1/2) = 0.125 I

0.

The reduction in intensity is by a factor of 0.875 of the incident intensity.

(b) By the same logic as in part (a) we have transmitted I0(1/2)(cos230°)(cos230°)(cos230°) =

I0(1/2)(cos230°)3 = 0.211 I

0. Then the fraction absorbed is 0.789 .

(c) Yet again we compute transmission I0(1/2)(cos215°)6 = 0.330 I

0. And the fraction absorbed

is 0.670 .

(d) We can get more and more of the incident light through the stack of ideal fi lters, approach-ing 50%, by reducing the angle between the transmission axes of each one and the next.

Additional Problems

*P38.42 The central bright fringe is wider than the side bright fringes, so the light must have been

diffracted by a single slit . For precision, we measure from the second minimum on one side

of the center to the second minimum on the other side:

2 11 7 6 3 5 4y y= − =( . . ) .cm = cm 22 7

0 027

2 60 010 4

0 595

.

tan.

..

.

cm

m

mθ

θ

= = =

= °

y

L==

=

= =× −

0 010 4

2 632 8 10 9

.

sin

sin

.

rad

m

a m

am

θ λ

λθ

(( )°

=×( )

= ×−

sin .

.

..

0 595

2 632 8 10

0 010 41 22 10

9 m −−4 m

P38.43 Let the fi rst sheet have its axis at angle q to the original plane of polarization, and let each further sheet have its axis turned by the same angle.

The fi rst sheet passes intensity Imax cos2 θ

The second sheet passes Imax cos4 θ

and the nth sheet lets through I Inmax maxcos .2 0 90θ ≥ where θ = °45

n

Try different integers to fi nd cos .2 5 45

50 885× °⎛

⎝⎜⎞⎠⎟ = cos .2 6 45

60 902× °⎛

⎝⎜⎞⎠⎟ =

(a) So n = 6

(b) θ = °7 50.


394 Chapter 38

P38.44 Consider vocal sound moving at 340 m/s and of frequency 3 000 Hz. Its wavelength is

λ = = =vf

340

3 0000 113

m s

Hzm.

If your mouth, for horizontal dispersion, behaves similarly to a slit 6.00 cm wide, then a msinθ λ= predicts no diffraction minima. You are a nearly isotropic source of this sound. It spreads out from you nearly equally in all directions. On the other hand, if you use a megaphone with width 60.0 cm at its wide end, then a msinθ λ= predicts the fi rst diffraction minimum at

θ λ= ⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =− −sin sin

.1 1 0 113m

a

m

0.600 m110 9. °

This suggests that the sound is radiated mostly toward the front into a diverging beam of angular diameter only about 20°. With less sound energy wasted in other directions, more is available for your intended auditors. We could check that a distant observer to the side or behind you receives less sound when a megaphone is used.

*P38.45 (a) We assume the fi rst side maximum is at a sinq = 1.5 l. (Its location is determined more precisely in problem 66.) Then the required fractional intensity is

I

I

a

amax

sin( sin / )

sin /

sin( .= ⎡⎣⎢

⎤⎦⎥ =π θ λ

π θ λπ

21 5 ))

. ..

1 5

1

2 250 0450

2

2π π⎡⎣⎢

⎤⎦⎥

= =

(b) Proceeding as in part (a) we have a sinq/l = 2.5 and

I

I

a

amax

sin( sin / )

sin /

sin( .= ⎡⎣⎢

⎤⎦⎥ =π θ λ

π θ λπ

22 5 ))

. ..

2 5

1

6 250 0162

2

2π π⎡⎣⎢

⎤⎦⎥

= =

*P38.46 The energy in the central maximum we can estimate in Figure 38.6 as proportional to

(height)(width) = Imax

(2p)

As in problem 45, the maximum height of the fi rst side maximum is approximately

Imax

[sin (3p /2)/(3p /2)]2 = 4Imax

/9p 2

Then the energy in one side maximum is proportional to p (4Imax

/9p 2),

and that in both of the fi rst side maxima together is proportional to 2p (4Imax

/9p 2).

Similarly and more precisely, and always with the same proportionality constant, the energy in both of the second side maxima is proportional to 2p (4I

max/25p 2).

The energy in all of the side maxima together is proportional to

2p (4Imax

/p 2)[1/32 + 1/52 + 1/72 + …] = 2p (4Imax

/p 2)[p 2/8 – 1] = (8Imax

/p )[p 2/8 – 1]

= Imax

(p – 8/p) = 0.595Imax

The ratio of the energy in the central maximum to the total energy is then

Imax

(2p)/{Imax

(2p) + 0.595Imax

} = 2p /6.88 = 0.913 = 91.3%

Our calculation is only a rough estimate, because the shape of the central maximum in particular is not just a vertically-stretched cycle of a cosine curve. It is slimmer than that.



P38.47 The fi rst minimum is at a sinθ λ= ( )1

This has no solution if λa

> 1

or if a < =λ 632 8. nm

*P38.48 With light in effect moving through vacuum, Rayleigh’s criterion limits the resolution according to

d/L = 1.22 l/D D = 1.22 lL/d = 1.22 (885 × 10–9 m) 12 000 m/2.3 m = 5.63 mm

The assumption is absurd. Over a horizontal path of 12 km in air, density variations associated with convection (“heat waves” or what an astronomer calls “seeing”) would make the motor-cycles completely unresolvable with any optical device.

P38.49 d = = ×−−1

4002 50 101

6

mmm.

(a) d msinθ λ= θa = × ××

⎛⎝⎜

⎞⎠⎟

= °−−

−sin .19

6

2 541 1025 6

m

2.50 10 m

(b) λ = × = ×−

−541 104 07 10

97m

1.33m. θb = × ×

×⎛⎝⎜

⎞⎠⎟

=−−

−sin.

.17

6

2 4 07 1019 0

m

2.50 10 m°°

(c) d asinθ λ= 2 and dnbsinθ λ= 2

combine by substitution to give n b asin sinθ θ= ( )1

P38.50 (a) λ = vf

: λ = ××

=−

3 00 10

1 40 100 214

8

9 1

.

..

m s

sm

θ λmin .= 1 22

D: θ µmin .

..=

×⎛⎝

⎞⎠ =1 22

0 2147 26

m

3.60 10 mrad4

θ µπmin . .= × ×⎛

⎝⎞⎠ =7 26

180 60 601 50rad

sarc secoonds

(b) θmin = d

L: d L= = ×( )( ) =−θmin . .7 26 10 26 000 0 1896 rad ly ly

(c) It is not true for humans, but we assume the hawk’s visual acuity is limited only by Rayleigh’s criterion.

θ λ

min .= 1 22D

θ µmin . .= ××

⎛⎝⎜

⎞⎠⎟

=−

−1 22500 10

50 89

3

m

12.0 10 mrrad seconds of arc10 5.( )

(d) d L= = ×( )( ) = ×− −θmin . . .50 8 10 30 0 1 52 106 3rad m m == 1 52. mm


396 Chapter 38

P38.51 With a grazing angle of 36.0°, the angle of incidence is 54.0°

tan tan . .θ p n= = ° =54 0 1 38

In the liquid, λ λn n

= = =750545

nm

1.38nm .

P38.52 (a) Bragg’s law applies to the space lattice of melanin rods. Consider the planes d = 0 25. mµ apart. For light at near-normal incidence, strong refl ection happens for the wavelength given by 2d msinθ λ= . The longest wavelength refl ected strongly corresponds to m = 1:

2 0 25 10 90 16. sin×( ) ° =− m λ λ = 500 nm. This is the blue-green color.

(b) For light incident at grazing angle 60°, 2d msinθ λ=

gives 1 2 0 25 10 60 4336λ = ×( ) ° =−. sinm nm. This is violet.

(c) Your two eyes receive light refl ected from the feather at different angles, so they receive light incident at different angles and containing different colors reinforced by constructive interference.

(d) The longest wavelength that can be refl ected with extra strength by these melanin rods is the one we computed fi rst, 500 nm blue-green.

(e) If the melanin rods were farther apart (say 0 32. mµ ) they could refl ect red with constructive interference.

P38.53 (a) d msinθ λ=

or dm= =

×( )°

=−λ

θµ

sin sin ..

3 500 10

32 02 83

9 mm

Therefore, lines per unit length = =× −

1 1

2 83 10 6d . m

or lines per unit length = × = ×− −3 53 10 3 53 105 1 3 1. .m cm .

(b) sin.

.θ λ= =×( )

×= ( )

−

−

m

d

mm

500 10

2 83 100 177

9

6

m

m

For sin .θ ≤ 1 00, we must have m 0 177 1 00. .( ) ≤

or m ≤ 5 66.

Therefore, the highest order observed is m = 5

Total number of primary maxima observed is 2 1 11m + =



*P38.54 (a) For the air-to-water interface,

tan.

.θ p

n

n= =water

air

1 33

1 00 θ p = °53 1.

and 1 00 1 33 2. sin . sin( ) = ( )θ θp

θ21 53 1

1 3336 9= °⎛

⎝⎞⎠ = °−sin

sin .

..

For the water-to-slab interface, tan tan.

θ θp

n

n

n= = =3 1 33slab

water

so θ31 1 33= −tan ( / . )n .

The angle between surfaces is θ θ θ= − =3 2 tan−1(n /1.33) − 36.9°.

(b) If we imagine n → ∞, then θ → 53.1°. The material of the slab in this case is higher in opti-cal density than any gem. The light in the water skims along the upper surface of the slab.

(c) If we imagine n = 1, then θ = 0. The slab is so low in optical density that it is like air. The light strikes parallel surfaces as it enters and exits the water, both at the polarizing angle.

P38.55 A central maximum and side maxima in seven orders of interference appear. If the seventh order is just at 90°,

d msinθ λ= d 1 7 654 10 9( ) = ×( )− m d = 4 58. mµ

If the seventh order is at less than 90°, the eighth order might be nearly ready to appear according to

d 1 8 654 10 9( ) = ×( )− m d = 5 23. mµ

Thus 4 58 5 23. .m mµ µ< <d .

P38.56 (a) We require θ λmin .= = =1 22

2D L

D

L

radius of diffraction disk

Then D L2 2 44= . λ

(b) D = ×( )( ) =−2 44 500 10 0 150 4289. .m m mµ

P38.57 For the limiting angle of resolution between lines we

assume θ λmin . .

.= =

×( )×( ) =

−

−1 22 1 22550 10

5 00 101

9

3D

m

m..34 10 4× − rad.

Assuming a picture screen with vertical dimension �, the minimum viewing distance for no

visible lines is found from θmin = � 485

L. The desired ratio is then

L

�= =

×( ) =−

1

485

1

485 1 34 1015 44θmin .

.rad

When the pupil of a human eye is wide open, its actual resolving power is signifi cantly poorer than Rayleigh’s criterion suggests.

FIG. P38.55

FIG. P38.54

AirWaterq3q2

qp

q


398 Chapter 38

P38.58 (a) Applying Snell’s law gives n n2 1sin sinφ θ= . From the sketch, we also see that:

θ φ β π+ + = , or φ π θ β= − +( ) Using the given identity:

sin sin cos cos sinφ π θ β π θ β= +( ) − +( ) which reduces to: sin sinφ θ β= +( ) Applying the identity again: sin sin cos cos sinφ θ β θ β= +

Snell’s law then becomes: n n2 1sin cos cos sin sinθ β θ β θ+( ) =

or (after dividing by cosq ): n n2 1tan cos sin tanθ β β θ+( ) =

Solving for tanθ gives: tansin

cosθ β

β=

−n

n n2

1 2

(b) If β = °90 0. , n1 1 00= . , and n n2 = , the above result becomes:

tan.

.θ = ( )

−n 1 00

1 00 0, or n = tanθ , which is Brewster’s law.

P38.59 (a) From Equation 38.1, θ λ= ⎛⎝⎜

⎞⎠⎟

−sin 1 m

a

In this case m = 1 and λ = = ××

= × −c

f

3 00 10

7 50 104 00 10

8

92.

..

m s

Hzm

Thus, θ = ××

⎛⎝⎜

⎞⎠⎟

= °−−

−sin.

..1

2

2

4 00 10

6 00 1041 8

m

m

(b) From Equation 38.2, I

Imax

sin= ( )⎡

⎣⎢

⎤

⎦⎥

φφ

2

where φ π θλ

= a sin

When θ = °15 0. , φπ

=( ) °

=0 060 0 15 0

0 040 01 22

. sin .

..

m

mrad

and I

Imax

sin .

..= ( )⎡

⎣⎢⎤⎦⎥

=1 22

1 220 593

2rad

rad

(c) sinθ λ=a

so θ = °41 8. :

This is the minimum angle subtended by the two sources at the slit. Let a be the half angle between the sources, each a distance � = 0 100. m from the center line and a distance L from the slit plane. Then,

L = = ( ) °⎛⎝

⎞⎠ =� cot . cot

..α 0 100

41 8

20 262m m

FIG. P38.59(c)

FIG. P38.58(a)



*P38.60 (a) The fi rst sheet transmits one-half the intensity of the originally unpolarized light, because the average value of the cosine-squared function in Malus’s law is one-half. Then

I

Imax

cos . cos .= °)( °( ) =1

245 0 45 0

1

82 2

(b) No recipes remain. The two experiments follow precisely analogous steps, but the results are different. The middle fi lter in part (a) changes the polarization state of the light that passes through it, but the recipe selections do not change individual recipes. The result for light gives us a glimpse of how quantum-mechanical measurements differ from classical measurements.

P38.61 (a) Constructive interference of light of wavelength l on the screen is described by

d msinθ λ= where tanθ = y

L so sinθ =

+y

L y2 2. Then d y L y m( ) +( ) =

−2 2 1 2 λ .

Differentiating with respect to y gives

d L y d y L y y md

11

20 22 2 1 2 2 2 3 2

+( ) + ( ) −⎛⎝

⎞⎠ +( ) +( ) =

− − λλ

λdy

d

L y

d y

L ym

d

dy

d L d2 2 1 2

2

2 2 3 2

2

+( ) − ( )+( ) = = ( ) + (( ) − ( )

+( )= ( )

+( )

y d y

L y

d

dy

d L

m L y

2 2

2 2 3 2

2

2 2 3 2

λ

(b) Here d msinθ λ= gives 10

8 0001 550 10

29

−−= ×( )m

msinθ , θ = ××

⎛⎝⎜

⎞⎠⎟

= °−−

−sin.

.16

6

0 55 10

1026 1

m

1.25 m

y L= = ° =tan . tan . .θ 2 40 26 1 1 18m m

Now d

dy

dL

m L y

λ =+( ) = × ( )−2

2 2 3 2

6 21 25 10

1 2 4

.

.

m 2.40 m

m mnm cm

( ) + ( )( )= × =−

2 2 3 27

1 183 77 10 3 77

.. . .

P38.62 (a) The angles of bright beams diffracted from the grating are given by d m( ) =sinθ λ. The

angular dispersion is defi ned as the derivative d

d

θλ

: dd

dm( ) =cosθ θ

λ d

d

m

d

θλ θ

=cos

(b) For the average wavelength 578 nm,

d msinθ λ= 0 02

2 578 10 9.sin

m

8 000mθ = ×( )−

θ = × ××

= °−−

−sin .19

6

2 578 1027 5

m

2.5 10 m

The separation angle between the lines is

∆ ∆ ∆θ θλ

λθ

λ= = =× °

×−

d

d

m

d cos . cos ..

2

2 5 10 27 52 116 m

110

0 001 90 0 001 90 0 001 90180

9−

= = =

m

rad rad. . .°°⎛

⎝⎞⎠ = °

π rad0 109.


400 Chapter 38

P38.63 (a) The E and O rays, in phase at the surface of the plate, will have a phase difference

θ πλ

δ= ⎛⎝

⎞⎠

2

after traveling distance d through the plate. Here d is the difference in the optical path lengths of these rays. The optical path length between two points is the product of the actual path length d and the index of refraction. Therefore,

δ = −dn dnO E

The absolute value is used since n

nO

E

may be more or less than unity. Therefore,

θ πλ

πλ

= ⎛⎝

⎞⎠ − = ⎛

⎝⎞⎠ −2 2

dn dn d n nO E O E

(b) dn nO E

=−

=×( )( )

−=

−λθπ

ππ2

550 10 2

2 1 544 1 5531

9 m

. ... .53 10 15 35× =− m mµ

P38.64 (a) From Equation 38.2, I

Imax

sin= ( )⎡

⎣⎢

⎤

⎦⎥

φφ

2

where we defi ne φ π θλ

≡ a sin

Therefore, when I

Imax

= 1

2 we must have

sinφφ

= 1

2, or sinφ φ=

2

(b) Let y1 = sinφ and y2 2= φ

.

A plot of y1 and y2 in the range 1 002

. ≤ ≤φ π is shown to

the right.

The solution to the transcendental equation is found to

be φ = 1 39. rad .

(c) π θ

λφa sin = gives sin .θ φ

πλ λ= ⎛

⎝⎞⎠ =

a a0 443 .

If λa

is small, then θ λ≈ 0 443.a

.

This gives the half-width, measured away from the maximum at θ = 0. The pattern is symmetric, so the full width is given by

∆θ λ λ λ= − −⎛⎝⎜

⎞⎠⎟ =0 443 0 443

0 886. .

.

a a a

FIG. P38.64(b)



P38.65 f 2 sinφ 1 1.19 bigger than f2 1.29 smaller than f1.5 1.41 smaller

1.4 1.394

1.39 1.391 bigger

1.395 1.392

1.392 1.391 7 smaller

1.391 5 1.391 54 bigger

1.391 52 1.391 55 bigger

1.391 6 1.391 568 smaller

1.391 58 1.391 563

1.391 57 1.391 561

1.391 56 1.391 558

1.391 559 1.391 557 8

1.391 558 1.391 557 5

1.391 557 1.391 557 3

1.391 557 4 1.391 557 4

We get the answer as 1.391 557 4 to seven digits after 17 steps. Clever guessing, like using the value of 2 sinφ as the next guess for f, could reduce this to around 13 steps.

P38.66 In I I= ( )⎡

⎣⎢

⎤

⎦⎥max

sin φφ

2

we fi nd dI

dI

φφ

φφ φ φ

φ= ( )⎛

⎝⎜⎞⎠⎟

( ) ( ) − ( )( )max

sin cos sin22

⎡⎡

⎣⎢⎢

⎤

⎦⎥⎥

and require that it be zero. The possibility sin φ( ) = 0 locates all of the minima and the central maximum, according to

φ π π= 0 2, , , . . .; φ π θλ

π π= =a sin, , , . . .0 2 ; a sin , , , . . .θ λ λ= 0 2

The side maxima are found from φ φ φcos sin( ) − ( ) = 0, or tan φ φ( ) =

This has solutions φ = 4 493 4. , φ = 7 725 3. , and others, giving

(a) π θ λa sin .= 4 493 4 a sin .θ λ= 1 430 3

(b) π θ λa sin .= 7 725 3 a sin .θ λ= 2 459 0


402 Chapter 38

P38.67 The fi rst minimum in the single-slit diffraction pattern occurs at

sin minθ λ= ≈a

y

L

Thus, the slit width is given by

aL

y= λ

min

For a minimum located at ymin . .= ±6 36 0 08mm mm, the width is

a =×( )( )

×=

−

−

632 8 10 1 00

6 36 1099 5

9

3

. .

..

m m

mmµ ±±1%

ANSWERS TO EVEN PROBLEMS

P38.2 547 nm

P38.4 91 2. cm

P38.6 (a) 51.8 mm wide and 949 mm high (b) horizontal; vertical. See the solution. A smaller distance between aperture edges causes a wider diffraction angle. The longer dimension of each rectangle is 18.3 times larger than the smaller dimension.

P38.8 See the solution.

P38.10 (a) 0°, 10 3. °, 21 0. °, 32 5. °, 45 8. °, 63 6. ° (b) nine bright fringes, at 0° and on either side at 10 3. °, 21 0. °, 32 5. °, and 63 6. ° (c) 1 00. , 0 811. , 0 405. , 0 090 1. , 0 032 4.

P38.12 (a) 79.8 mrad (b) violet, 54.2 mrad (c) The resolving power is improved, with the minimum resolvable angle becoming 60.0 mrad.

P38.14 Between 186 m and 271 m; blue is resolvable at larger distances because it has shorter wavelength.

P38.16 30 5. m

P38.18 105 m

P38.20 514 nm

P38.22 1 81. mµ


P38.26 74.2 grooves/mm

P38.28 (a) 0 738. mm (b) See the solution.

P38.30 93.4 pm

P38.32 5.51 m, 2.76 m, 1.84 m

FIG. P38.67



P38.34 3

8

P38.36 1.11



P38.42 One slit 0.122 mm wide. The central maximum is twice as wide as the other maxima.



P38.48 5.63 mm. The assumption is absurd. Over a horizontal path of 12 km in air, density variations associated with convection (“heat waves” or what an astronomer calls “seeing”) would make the motorcycles completely unresolvable with any optical device.

P38.50 (a) 1.50 sec (b) 0.189 ly (c) 10.5 sec (d) 1.52 mm


P38.54 (a) q = tan−1(n/1.33) − 36.9° (b) If we imagine n → ∞, then q → 53.1°. The material of the slab in this case is higher in optical density than any gem. The light in the water skims along the upper surface of the slab. (c) If we imagine n = 1, then q = 0. The slab is so low in optical density that it is like air. The light strikes parallel surfaces as it enters and exits the water, both at the polarizing angle.

P38.56 (a) See the solution. (b) 428 mµ


P38.60 (a) 1/8 (b) No recipes remain. The two experiments follow precisely analogous steps, but the results are different. The middle fi lter in part (a) changes the polarization state of the light that passes through it, but the recipe selection processes do not change individual recipes.

P38.62 (a) See the solution. (b) 0.109°

P38.64 (a) See the solution. (b) and (c) See the solution.

P38.66 (a) a sin .θ λ= 1 430 3 (b) a sin .θ λ= 2 459 0


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