PS5Sol_f10 - Physics 112 Fall 2010 Professor William...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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 /
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern