PS5Sol_f10

# PS5Sol_f10 - Physics 112 Fall 2010 Professor William...

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Physics 112 Fall 2010 Professor William Holzapfel Homework 5 Solutions Problem 1: Seeing in the dark (a) Treating the filament as a blackbody, we know that total power is given by P b = σ B T 4 × A , so the area is: A = 100 W σ B T 4 = 100 . 30 * 5 . 67 * 10 - 8 * (3500) 4 m 2 = 3 . 9 * 10 - 5 m 2 (b) Treating your eye as a box of photons at temperature T b 300 K , the flux incident on your retina from the inside of your eye is Flux b = σ b T 4 b . The power from the filament spreads out over a sphere of radius D , the distance from the filament to your eye, so Flux f = σ b T 4 f A 4 πd 2 . Thus the ratio of the fluxes is Flux b Flux f = σ b T 4 b σ b T 4 f A 4 πd 2 = 300 K 3500 K 4 4 π (10 m) 2 0 . 3 * 3 . 9 * 10 - 5 m 2 5 , 800 . The reason we can see anything with so much background flux on our retina is our retina is only sensitive to a narrow band of wavelengths in the visible spectrum, and the background flux from our body peaks in the far infrared, contributing very little flux in the visible spectrum. (c) To figure out how far away we can see an object, we need to consider the flux from the object within the visible spectrum as compared to that of our eye. We can assume the visible band is so narrow we can treat it as infinitesimal v . Instead of numerically integrating the specific flux over the visible range, we’ll assume we can just take the total flux in the visible to be F ν ( ω v ) v , where F ν ( ω v ) is the specific flux at the middle of the visible spectrum. The total flux in the visible spectrum from the eye is then Flux b = F ν ( ω v ) = π 2 c 2 ω 3 v e ω v b - 1 and the total flux from the filament is: Flux f = F ν ( ω v ) = π 2 c 2 ω 3 v e ω v f - 1 × A 4 πD 2 . In the limit that we can just barely see the filament these two fluxes should be equal, which gives the relation D = A 4 π e ω v b - 1 e ω v f - 1 . Now we’re told to assume that the peak frequency of the filament is in the visible band, so we can find ω v by using Wein’s displacement law ω v k B T f = 2 . 82 1

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(for the purposes of this problem we don’t need to worry about how we’re defining ‘peak’ frequency like we did in problem set 4). Plugging this in the above expression, and using the fact that e 2 . 82 τ f b 1 we have: D = A 4 π e 2 . 82 τ f τ b - 1 e 2 . 82 - 1 A 4 π e 2 . 82 τ f τ b e 2 . 82 - 1 . If we plug in for T f and T b we get D 3200m = 3 . 2km. Maybe on a dark night out at sea you could see a lightbulb two miles away... Problem 2: Liquid 4 He (a) It’s tempting to just plug in the the giving quantities into the expression for Debye temperature that we found in lecture. However, looking back at that calculation we realize that we assumed that there were 3 possible polarizations of of phonons; in the case of liquid 4 He we’re told that there are no transverse phonons, so there is only one polarization (longitudinal) per mode. Looking back at how we derive the Debye temperature, we see that the total number of phonon modes must add up to 3 N . So with one polarization, the expression for the cutoff mode n max becomes: 1 8 n max 0 4 πn 2 dn = 3 N n max = 18 N π 1 /
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