University Physics Volume 3

# Chapter 4

### Check Your Understanding

4.1

$17.8°17.8°$, $37.7°37.7°$, $66.4°66.4°$; no

4.2

$74.3°74.3°$, $0.0083I00.0083I0$

4.3

From $dsinθ=mλdsinθ=mλ$, the interference maximum occurs at $2.87°2.87°$ for $m=20.m=20.$ From Equation 4.1, this is also the angle for the second diffraction minimum. (Note: Both equations use the index m but they refer to separate phenomena.)

4.4

$3.332×10−6m3.332×10−6m$ or 300 lines per millimeter

4.5

$8.4×10−4rad8.4×10−4rad$, 3000 times broader than the Hubble Telescope

4.6

$38.4°38.4°$ and $68.8°68.8°$; Between $θ=0°→90°θ=0°→90°$, orders 1, 2, and 3, are all that exist.

### Conceptual Questions

1 .

The diffraction pattern becomes wider.

3 .

Walkie-talkies use radio waves whose wavelengths are comparable to the size of the hill and are thus able to diffract around the hill. Visible wavelengths of the flashlight travel as rays at this size scale.

5 .

The diffraction pattern becomes two-dimensional, with main fringes, which are now spots, running in perpendicular directions and fainter spots in intermediate directions.

7 .

The parameter $β=ϕ/2β=ϕ/2$ is the arc angle shown in the phasor diagram in Figure 4.7. The phase difference between the first and last Huygens wavelet across the single slit is $2β2β$ and is related to the curvature of the arc that forms the resultant phasor that determines the light intensity.

9 .

blue; The shorter wavelength of blue light results in a smaller angle for diffraction limit.

11 .

No, these distances are three orders of magnitude smaller than the wavelength of visible light, so visible light makes a poor probe for atoms.

13 .

UV wavelengths are much larger than lattice spacings in crystals such that there is no diffraction. The Bragg equation implies a value for sin⁡θ greater than unity, which has no solution.

15 .

Image will appear at slightly different location and/or size when viewed using $10%10%$ shorter wavelength but at exactly half the wavelength, a higher-order interference reconstructs the original image, different color.

### Problems

17 .

a. $33.4°33.4°$; b. no

19 .

a. $1.35×10−6m1.35×10−6m$; b. $69.9°69.9°$

21 .

750 nm

23 .

2.4 mm, 4.7 mm

25 .

a. $1.00λ;1.00λ;$ b. $50.0λ;50.0λ;$ c. $1000λ1000λ$

27 .

1.92 m

29 .

$45.1 ° 45.1 °$

31 .

$I / I 0 = 2.2 × 10 −5 I / I 0 = 2.2 × 10 −5$

33 .

$0.63 I 0 , 0.11 I 0 , 0.0067 I 0 , 0.0062 I 0 , 0.00088 I 0 0.63 I 0 , 0.11 I 0 , 0.0067 I 0 , 0.0062 I 0 , 0.00088 I 0$

35 .

0.200

37 .

3

39 .

9

41 .

$5.97 ° 5.97 °$

43 .

$8.99 × 10 3 8.99 × 10 3$

45 .

707 nm

47 .

a. $11.8°11.8°$, $12.5°12.5°$, $14.1°14.1°$, $19.2°19.2°$; b. $24.2°24.2°$, $25.7°25.7°$, $29.1°29.1°$, $41.0°41.0°$; c. Decreasing the number of lines per centimeter by a factor of x means that the angle for the x-order maximum is the same as the original angle for the first-order maximum.

49 .

a. using $λ=700nm,θ=5.0°;λ=700nm,θ=5.0°;$ b. using $λ=460nm,θ=3.3°λ=460nm,θ=3.3°$

51 .

a. 26,300 lines/cm; b. yes; c. no

53 .

$1.13 × 10 −2 m 1.13 × 10 −2 m$

55 .

107 m

57 .

a. $7.72×10−4rad;7.72×10−4rad;$ b. 23.2 m; c. 590 km

59 .

a. $2.24×10−4rad;2.24×10−4rad;$ b. 5.81 km; c. 0.179 mm; d. can resolve details 0.2 mm apart at arm’s length

61 .

$2.9 μ m 2.9 μ m$

63 .

6.0 cm

65 .

7.71 km

67 .

1.0 m

69 .

1.2 cm or closer

71 .

no

73 .

0.120 nm

75 .

$4.51 ° 4.51 °$

77 .

$13.2 ° 13.2 °$

79 .

a. 2.2 mm; b. $0.172°0.172°$, second-order yellow and third-order violet coincide

81 .

2.2 km

83 .

1.3 cm

85 .

a. 0.28 mm; b. 0.28 m; c. 280 m; d. 113 km

87 .

33 m

89 .

a. vertically; b. $±20°±20°$, $±44°±44°$; c. 0, $±31°±31°$, $±60°±60°$; d. 89 cm; e. 71 cm

91 .

0.98 cm

93 .

$I / I 0 = 0.041 I / I 0 = 0.041$

95 .

340 nm

97 .

a. 0.082 rad and 0.087 rad; b. 480 nm and 660 nm

99 .

two orders

101 .

yes and N/A

103 .

600 nm

105 .

a. $3.4×10−5°3.4×10−5°$; b. $51°51°$

107 .

0.63 m

109 .

1

111 .

0.17 $mW/cm2mW/cm2$ for $m=1m=1$ only, no higher orders

113 .

$28.7 ° 28.7 °$

115 .

a. 42.3 nm; b. This wavelength is not in the visible spectrum. c. The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light.

117 .

a. 549 km; b. This is an unreasonably large telescope. c. Unreasonable to assume diffraction limit for optical telescopes unless in space due to atmospheric effects.

### Challenge Problems

119 .

a. $I=0.00500I0,0.00335I0I=0.00500I0,0.00335I0$; b. $I=0.00500I0,0.00335I0I=0.00500I0,0.00335I0$

121 .

12,800

123 .

$1.58 × 10 −6 m 1.58 × 10 −6 m$

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