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Radiation Pressure problem

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Estimate the radiation pressure due to a 100-W bulb at a distance of 5.0 cm from the center of the bulb. Assume that light is complitely absorbed.

    Estimate the force exerted on your fingertip if you place it at this point. Assume area of the fingertip to be 1.5 cm2.

    2. Relevant equations

    P = I/c
    I = Power/(4*pi*d2)??

    3. The attempt at a solution
    I don't know how to solve for intensity. I think it should have units of W/m^2 so I would guess it would be 100/(4*pi*.05*.05) = Power/(4*pi*d2), since 4*pi*d*d would be the area of a sphere of radius d but I doubt it is correct.

    For the second part it would just be the pressure I get in the first part times 0.00015 (1.5 cm2 in m2) for force.

    Thanks in advance!
     
  2. jcsd
  3. Mar 30, 2009 #2

    mgb_phys

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    Yes, you need the surface area of a sphere radius=5cm.
    Then you have the power/area at that distance ( since all the power from the bulb reaches the surface of this sphere)
    Now you need to know how many photons/sec from the bulb and how much momentum/photon.

    It's probably easier to do this in stages rather than just write a single equation - be carefull of the units.
     
  4. Mar 30, 2009 #3
    So am I correct in saying that pressure is intensity over the speed of light? If so, am I also correct in my formula for intensity?

    I'm not quite sure how I work in momentum/second (momentum/photon and photons/second).
     
  5. Mar 30, 2009 #4

    mgb_phys

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    The momentum of a photon is h / wavelength - where h is Planks constant.
    The energy of a photon is h * frequency or h * speed of light / wavelength.

    You will have to estimate an average wavelength for the light bulb (hint the filament is around 2-3000K + Wein's law)
     
  6. Mar 30, 2009 #5
    How is this not giving me more than I need? The title's radiation is just about light not radiation per se.

    My units work out if I have (100 W/(4*pi*.05*.05 m2)/c = 1.something*10-5 pascals which are the units I want for the question (thank you masteringphysics)
     
  7. Mar 30, 2009 #6

    mgb_phys

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    100/(4*pi*.05*.05) is the intensity, ie the power/area at that distance.
    To go from power to momentum you need to work out the number of photons and the momentum of each.
     
  8. Mar 30, 2009 #7
    Why can't I simply divide intensity by the speed of light which would make my units work out?

    I submitted my solutions; Pressure = [tex]/frac{/frac{100}{4/pi .05^2}}{c}[/tex] worked.
     
    Last edited: Mar 30, 2009
  9. Mar 30, 2009 #8
    That should be [TEX]\frac{\frac{100 W}{4pi .05^2 m^2}}{c}[/TEX]

    And for some reason this is coming out as mu_0*8/(2*pi*(.0447213595)) for some reason...
     
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