Ch. 25 HW Solutions
2.) Given the wavelength, λ, the frequency is found using c = λf or f = c/λ = 3×108 /633×
10−9 = 4.739 × 1014 Hz.
3.) Given the wavelength in vacuum and knowing the speed of light in vacuum, one can
calculate the frequency in vacuum, f = c/λ = 5.09 × 1014 Hz. The frequency of a wave in
water will be the same as the frequency in vacuum. (Imagine a wave incident on the water.
The E-M wave at the point in space where the water begins will have the frequency of the
wave in vacuum, and thus the same frequency will propagate into the water. The wavelengths in the different media will change as a result of the speed of light being different in
different media; this will be discussed in greater detail in later chapters.) Since the frequency
is the same in both vacuum and water, we can write cv = λv f and cw = λw f where v and
w represent vacuum and water respectively. Diving the second equation by the first yields
cw /cv = λw /λv . Thus, λw = 2.25/3.00λv = 442 nm.
7.) (a) If the circuit resonates to 2.0 cm microwaves, its resonant
√ frequency will be given
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by fo = c/λ = 3 × 10 /0.02 = 1.5 × 10 Hz. (b) fo = 1/2π LC or C = 1/(2πfo )2 L =
2.253 × 10−19 F
13.) P = 3.0 × 10−3 W, d = 2.0 × 10−3 m, so r = 1.0 × 10−3 m. Intensity is defined
as power per unit area, I = P/A = P/πr2 = 3.0 × 10−3 /π(1.0 × 10−3 )2 = 954.93 W/m2 .
14.) I = 1340 W/m2 from the sun incident on area A = 20 m2 . The total power striking the solar panel is Ptot = IA = (1340)(20) = 26800 W. The panel is only 8% efficient
at converting solar power into useable energy, so the effective power the panel provides is
Pef f = (0.08)Ptot = 2144 W. At 60 W per bulb, the number of bulbs that can be lit is
Pef f /Pbulb = 2144/60 = 35.733. The panel can light 35 bulbs.
17.) Luminous intensity is I = Φ/Ω = 40 cd. For a full sphere, Ω = 4π. So the spherical flux is Φ = IΩ = (40)(4π) = 502.7 lm.
18.) Luminous intensity is I = Φ/Ω = 1500/4π = 119.4 cd for a full sphere.
19.) Solid (spherical) angle is given by Ω = A/R2 . Since A = π(d/2)2 = 3.14 × 10−6 m2 and
the distance from source to eye is R = 1.5 m, Ω = A/R2 = 3.14 × 10−6 /1.52 = 1.40 × 10−6
sr. A very small spherical angle with respect to a full spere (4π sr).
20.) The area of the paper is 0.2 × 0.25 = 0.05 m2 . The distance from source is 1.80
m, so Ω = A/R2 = 0.05/1.82 = 0.0154 sr. This is only approximate because the area in the
solid angle equation actually describes the area of the curved surface of a sphere, not a flat
surface like the paper. The smaller the area or farther from the source it is, the equation
becomes more accurate.
21.) Φ = IΩ = (40)(4π) = 502.6 lm.
23.) Φ = IΩ = (130)(4π) = 1633.6 lm.
24.) At Earth, the Sun’s intensity = 1340 W/m2 . At the distance of the earth from the sun,
R = 1.5×1011 m, all the power emitted from the sun passes through a sphere with this radius
with the intensity given; it’s area is A = 4πR2 = 4π(1.5 × 1011 )2 = 2.83 × 1023 m2 . Thus, the
total power emitted from the sun is given by P =(intensity)(area)= (1340)(2.83 × 1023 ) =
3.79 × 1026 W. Since [W] = [J/s], the energy emitted from the sun each hour (3600 seconds)
is E = P t = (3.79 × 1026 J/s)(3600 s) = 1.364 × 1030 J per hour.
25.) From the previous problem, the Sun’s power is given by Psun = 3.79 × 1026 W. The distance of Mercury from the sun is Rmerc = 5.8 × 1010 m. The intensity of the Sun’s radiation
at this distance is given by its total power divided by the area of the sphere that it crosses
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at this radius, (intensity) = Psun /A = Psun /4πRmerc
= 8962.54 W/m2 . Much higher than
the intensity of the sun at Earth. Mercury is hot.
26.) Same as previous problem but the distance is now Rpluto = 5.9 × 1012 m. The intensity at this distance is 0.866 W/m2 . Much less than that of Mercury or Earth. Pluto is
cold.
27.) = Φ/P = 1170 lm/75 W= 15.6 lm/W.
28.) Φ = P = (15)(60) = 900 lm.
29.) A = 600 cm2 = 0.06 m2 , Φ = 3000 lm. E = Φ/A = 3000/0.06 = 50000 lux (or
lm/m2 ).
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