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Force of a Laser on a Mirror

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    A collimated laser beam of power P = 106 Watt illuminates a 100%-reflecting
    micromirror at an angle of incidence of θ=30 for duration of = 0.015 seconds. Prior to laser
    illumination, the mirror is at rest. What is the mass of the mirror m if it acquires velocity
    of 1 m/s after laser illumination?
     
  2. jcsd
  3. Mar 30, 2013 #2
    I'd take a guess that you need to find a formula to do with the relationship between energy and momentum for light.

    Do you have any more information about the problem sheet you're working on?
     
  4. Mar 30, 2013 #3

    Astronuc

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    Staff: Mentor

    Please write the relevant equations and demonstrate effort by showing one's attempt as solving the problem.

    Note that photons have energy and momentum.
     
  5. Apr 6, 2013 #4
    Here is my cut at it. (Had some family issues, sorry for the delay)

    1. The problem statement, all variables and given/known data

    A collimated laser beam of power P = 106 Watt illuminates a 100%-reflecting
    micromirror at an angle of incidence of ?=30 for duration of = 0.015 seconds. Prior to laser
    illumination, the mirror is at rest. What is the mass of the mirror m if it acquires velocity
    of 1 m/s after laser illumination?

    Power = 106 (W)
    Angle of Incidence = 30 degrees
    Time = 0.015 (s)
    Velocity = (1 m/s)
    Mass = ?

    2. Relevant equations

    No equations were given.

    Force = Mass*Acceleration

    Force = 2*(Intensity*Area)/c
    = 2*Power/c

    (There is twice the force because it is a totally reflected wave)

    3. The attempt at a solution

    Force = 2*(Intensity*Area)/c
    = 2*Power/c
    = 2*(106*cos(30) W)/(3*10^8 m/s)
    = 6.12*10^-7 N

    Force = Mass*Acceleration
    Mass = Force/Acceleration
    = (6.12*10^-7 N)/[(0 m/s-1 m/s)/(0.015 s)]
    = 9.18*10^-9 Kg


    It all seems too simple, so I would really appreciate a good review.
     
  6. Apr 6, 2013 #5

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is as simple as that. You can check your calculations with WolframAlpha or similar tools.
    The place where you add the angle is a bit strange - it influences the momentum transfer, and not the area. The result is the same, however.
     
  7. Apr 6, 2013 #6
    Thanks mfb.

    Would this be the correct way to account for the angle of incidence?

    Force = 2*(Intensity*Area)/c
    = 2*Power/c
    = 2*(106 W)/(3*10^8 m/s)
    = 7.07*10^-7 N

    Force = Mass*Acceleration
    Mass = Force/Acceleration
    = (7.07*10^-7*cos(30) N)/[(0 m/s-1 m/s)/(0.015 s)]
    = 9.18*10^-9 Kg
     
  8. Apr 7, 2013 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    I would change Force = 2*Power/c to Force = 2*cos(alpha)*Power/c. This way, it is clear where the angle comes from.
     
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