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Force of a Laser on a Mirror

  • Thread starter Newbie12
  • Start date
  • #1
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Homework Statement



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?
 

Answers and Replies

  • #2
137
4
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?
 
  • #3
Astronuc
Staff Emeritus
Science Advisor
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Homework Statement



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

Note that photons have energy and momentum.
 
  • #4
4
0
Here is my cut at it. (Had some family issues, sorry for the delay)

Homework Statement



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 = ?

Homework 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)

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.
 
  • #5
33,888
9,606
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.
 
  • #6
4
0
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
 
  • #7
33,888
9,606
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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