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College Physics for AP® Courses

# Chapter 27

1.

$1 / 1 . 333 = 0 . 750 1 / 1 . 333 = 0 . 750 size 12{1/1 "." "333"=0 "." "750"} {}$

3.

1.49, Polystyrene

5.

0.877 glass to water

6.

$0 . 516º 0 . 516º size 12{0 "." "516"°} {}$

8.

$1 . 22 × 10 − 6 m 1 . 22 × 10 − 6 m size 12{1 "." "22" times "10" rSup { size 8{ - 6} } m} {}$

10.

600 nm

12.

$2 . 06º 2 . 06º size 12{2 "." "06"°} {}$

14.

1200 nm (not visible)

16.

(a) 760 nm

(b) 1520 nm

18.

For small angles $sinθ−tanθ≈θ(in radians)sinθ−tanθ≈θ(in radians) size 12{"sin"θ - "tan"θ approx θ $$"in""radians"$$ } {}$.

For two adjacent fringes we have,

$dsinθm=mλdsinθm=mλ size 12{d"sin"θ rSub { size 8{m} } =mλ} {}$

and

$dsin θm+1=m+1λdsin θm+1=m+1λ size 12{d"sin "θ rSub { size 8{m+1} } = left (m+1 right )λ} {}$

Subtracting these equations gives

d sin θ m + 1 − sin θ m = m + 1 − m λ d θ m + 1 − θ m = λ tan θ m = y m x ≈ θ m ⇒ d y m + 1 x − y m x = λ d Δ y x = λ ⇒ Δ y = xλ d d sin θ m + 1 − sin θ m = m + 1 − m λ d θ m + 1 − θ m = λ tan θ m = y m x ≈ θ m ⇒ d y m + 1 x − y m x = λ d Δ y x = λ ⇒ Δ y = xλ d alignl { stack { size 12{d left ("sin"θ rSub { size 8{m+1} } - " sin"θ rSub { size 8{m} } right )= left [ left (m+1 right ) - m right ]λ} {} # d left (θ rSub { size 8{m+1} } - θ"" lSub { size 8{m} } right )=λ {} # "tan"θ rSub { size 8{m} } = { {y rSub { size 8{m} } } over {x} } approx θ"" lSub { size 8{m} } drarrow d left ( { {y rSub { size 8{m+1} } } over {x} } - { {y rSub { size 8{m} } } over {x} } right )=λ {} # d { {Δy} over {x} } =λ drarrow {underline {Δy= { {xλ} over {d} } }} {} } } {}

20.

450 nm

21.

$5 . 97º 5 . 97º size 12{5 "." "97"°} {}$

23.

$8 . 99 × 10 3 8 . 99 × 10 3 size 12{8 "." "99" times "10" rSup { size 8{3} } } {}$

25.

707 nm

27.

(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.

29.

589.1 nm and 589.6 nm

31.

$28.7º 28.7º size 12{"28" "." "69"°} {}$

33.

$43 . 2º 43 . 2º size 12{"43" "." 2°} {}$

35.

$90 . 0º 90 . 0º size 12{"90" "." 0°} {}$

37.

(a) The longest wavelength is 333.3 nm, which is not visible.

(b) 333 nm (UV)

(c) $6.58×103cm6.58×103cm size 12{6 "." "58" times "10" rSup { size 8{3} } "cm"} {}$

39.

$1 . 13 × 10 − 2 m 1 . 13 × 10 − 2 m size 12{1 "." "13" times "10" rSup { size 8{ - 2} } m} {}$

41.

(a) 42.3 nm

(b) Not a visible wavelength

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.

43.

(a) $33.4º33.4º size 12{"33" "." 4°} {}$

(b) No

45.

(a) $1.35×10−6m1.35×10−6m size 12{1 "." "35" times "10" rSup { size 8{ - 6} } m} {}$

(b) $69.9º69.9º size 12{"69" "." 9°} {}$

47.

750 nm

49.

(a) $9.04º9.04º size 12{9 "." "04"°} {}$

(b) 12

51.

(a) $0.0150º0.0150º size 12{0 "." "0150"°} {}$

(b) 0.262 mm

(c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable.

53.

(a) $30.1º30.1º size 12{"30" "." 1°} {}$

(b) $48.7º48.7º size 12{"48" "." 7°} {}$

(c) No

(d) $2θ1=(2)(14.5º)=29º,θ2−θ1=30.05º−14.5º=15.56º2θ1=(2)(14.5º)=29º,θ2−θ1=30.05º−14.5º=15.56º size 12{2θ rSub { size 8{1} } = $$2$$ $$"14" "." 5°$$ ="29"°,~θ rSub { size 8{2} } - θ rSub { size 8{1} } ="30" "." "05"° - "14" "." 5"°=""15" "." "56"°} {}$. Thus, $29º≈(2)(15.56º)=31.1º29º≈(2)(15.56º)=31.1º size 12{"29"° approx $$2$$ $$"15" "." "56"°$$ ="31" "." 1°} {}$.

55.

$23.6º23.6º size 12{"23" "." 6°} {}$ and $53.1º53.1º size 12{"53" "." 1°} {}$

57.

(a) $1.63×10−4rad1.63×10−4rad size 12{1 "." "63" times "10" rSup { size 8{ - 4} } "rad"} {}$

(b) 326 ly

59.

$1 . 46 × 10 − 5 rad 1 . 46 × 10 − 5 rad size 12{1 "." "46" times "10" rSup { size 8{ - 5} } "rad"} {}$

61.

(a) $3.04×10−7rad3.04×10−7rad size 12{3 "." "04" times "10" rSup { size 8{ - 7} } "rad"} {}$

(b) Diameter of $235 m235 m$

63.

5.15 cm

65.

(a) Yes. Should easily be able to discern.

(b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations.

70.

532 nm (green)

72.

83.9 nm

74.

620 nm (orange)

76.

380 nm

78.

33.9 nm

80.

$4 . 42 × 10 − 5 m 4 . 42 × 10 − 5 m size 12{4 "." "42" times "10" rSup { size 8{ - 5} } m} {}$

82.

The oil film will appear black, since the reflected light is not in the visible part of the spectrum.

84.

$45 . 0º 45 . 0º size 12{"45" "." 0°} {}$

86.

$45 . 7 mW/m 2 45 . 7 mW/m 2 size 12{"45" "." 7`"mW/m" rSup { size 8{2} } } {}$

88.

$90 . 0% 90 . 0% size 12{"90" "." 0%} {}$

90.

$I 0 I 0 size 12{I rSub { size 8{0} } } {}$

92.

$48 . 8º 48 . 8º size 12{"48" "." 8°} {}$

94.

$41 . 2º 41 . 2º size 12{"41" "." 2°} {}$

96.

(a) 1.92, not diamond (Zircon)

(b) $55.2º55.2º size 12{"55" "." 2°} {}$

98.

$B 2 = 0 . 707 B 1 B 2 = 0 . 707 B 1 size 12{B rSub { size 8{2} } =0 "." "707"B rSub { size 8{1} } } {}$

100.

(a) $2.07×10-22.07×10-2$ °C/s

(b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.

1.

(b)

3.

(b) and (c)

5.

(b)

7.

(b)

9.

(b)

11.

(d)

13.

(b)

15.

(d)

17.

(b)

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