# Force of a Laser on a Mirror

## 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?

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I'd take a guess that you need to find a formula to do with the relationship between energy and momentum for light.

Astronuc
Staff Emeritus

## 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.

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.

mfb
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.

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

mfb
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.