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Force of light on a mirror

  1. Oct 8, 2006 #1
    "A 4.5W flashlight with a wavelength of 550nm is shined on a perfect mirror that reflects all the light completely. a) How many photon emanate from the flashlight? b) What force is exerted on the mirror by the beam of light?"

    A is easy, 1.25x10^19 photon/sec.

    My first attempt tried to exploit momentum, but I kept getting a dt in there, so that my answer was never in N, but rather N/s.

    Then, after looking at units, I noticed that a Watt / (m/s) = N. So, I just said F = P / c ... so F = 4.5W / (3x10^8 m/s) = 1.5 x 10^-8 N.

    Does this make any sense? Atleast its a small force...
    Last edited: Oct 8, 2006
  2. jcsd
  3. Oct 8, 2006 #2


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    So you think the wavelength information is extraneous???

    You were on the right track with the momentum idea. What is the momentum of a 550nm photon? What is its energy? How many photons hit the mirror every second if the power is 4.5W? How much is their momentum changed when they "bounce off" the mirror? How is force related to a change in momentum over a time interval?
  4. Oct 8, 2006 #3
    Yea, I left out the first part of the question.

    Anyways, p = h / l (l = lambda), and F = dp/dt. Now, if it hits straight on, p1 = h / 550nm; and p2 = - h / 550nm and dp = 2 h / 550nm.

    Now, if I take dp and multiply by the answer from part a (photons/sec), I get 3x10^-8 N*photon. Photon isn't exactly a unit and can be dropped, leaving N.
  5. Oct 8, 2006 #4


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    You don't know how long it takes for an individual photon to be turned around at the mirror, and you don't need to know. What the problem is looking for is an average force on the mirror over a finite time interval. You can assume the time dt for each photon is the average time between photon hits on the mirror, and dp is the momentum change you found for each photon. If you know how many photons hit the mirror every second, you can calculate dt and you can calculate the average force. The units will be fine.
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