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Black Body Radiation at a Distance from Emitter

  1. Feb 18, 2014 #1
    1.
    6. A 100 W light bulb is designed to operate with it’s filament at 2000K. If the filament
    is a perfect cylindrical Black Body and 2 cm long,

    i) What must it’s diameter be (3 marks)

    ii) What will be the wavelength of the intensity peak in its emission. (2 marks)

    iii) You look at this bulb, from a distance of 1 m, through a pair of rose-tinted glasses
    that pass only wavelengths 630-650 nm. Assuming that the filament looks like a
    uniformly emitting rectangle that is 2cm long, and as wide as your answer to i)
    Estimate the optical power entering your pupil if it is 4 mm diameter.




    2. Stefan's Law : P=σAT^4 , Wiens Law : λT=2.879x10^-3



    3.For part 1 and 2 I have arrived to answers , i.) 1.68x10^-3 (using Stefans Law and rearranging the surface area of the cylinder) and ii.) 1.4485x10^-6 using Wiens Law

    I am confused on part iii. I believe what the process should b is , 1.) Calculate new power as it is no longer a cylinder and is now treated as a rectangle. 2.) Calculate the factor of which the power drops over the distance you are from it . 3.) Then calculate as a proportion of the total area ? The amount of power the pupil itself is receiving . I am not sure how to calculate the power change over the distance . I have drawn a diagram and I think that the light should emit in a circle (as we are only talking 2 dimensions ) Therefore use 1/r^2 law to calculate a proportion ? I'm not entirely sure.

    Any help would be appreciated ,
    Thanks
    Dan !
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 18, 2014 #2

    BvU

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    No 2D problem, but then: the 1/r2 you mention is a 3D outcome, so applicable (by happenstance?)

    Just to help me find my own calculation error, could you show the part i) calculation ? (I end up with 1.75 mm, which I also consider pretty thick)

    Also, Wien has 2.898 10-3 as a constant; where does the 2.879 come from ?

    Then: for iii you need a ##{dR\over d\lambda }##. Which one do you add under 2. relevant formulas ?
     
    Last edited: Feb 18, 2014
  4. Feb 18, 2014 #3
    Sorry my mistake I mean 2.898 .

    It may have something to do with Planck distribution , calculate B then multiply by some other values to get the right dimensions .. i.e. area , solid angle and frequency.

    In response to the diameter question in i.) -->
    P=σAT^4

    100= 5.67x10^-8 x Surface Area of the Cylinder x (2000)^4

    Rearrange to get Surface Area = 100/(5.67x10^-8)x(2000)^4

    Now looking at the Surface Area , we know there are two discs and the surface area of the "rectangle" which all together form to make the cylinder . Therefore surface area = (2)(∏d^2)/4 + ∏d(0.02)

    Rearranging this you end up with : d^2 +0.04d - 200/(∏x5.67x10^-8)x(2000)^4 = 0

    Using the Quadratic Formula you arrive to a positive and negative answer , disregarding the negative answer as you are looking for a length , I get 1.68x10^-3

    If I have made a mistake please let me know .
     
  5. Feb 18, 2014 #4

    BvU

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    No, I was the one who ignored the end discs (I am pretty sloppy, and after all, there has to be an entrance and an exit for the current...) which explains the ifference. So I don't think you made a mistake.

    You plans look good to me!
     
  6. Feb 18, 2014 #5

    SteamKing

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    As always, include units in your calculations and results.
     
  7. Feb 18, 2014 #6
    I managed to do this question via the following , calculating B , Then doing the following Power = B*Area of Emitter*Solid Angle of the Pupil* Range of frequencies.

    Thanks guys !
     
  8. Feb 15, 2015 #7
    You are totally correct.
     
    Last edited: Feb 15, 2015
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