Flash of a laser moves a mirror hanging on a rope

[caspar]

Hallo to all members, this is my first post.

Homework Statement

A short Laser flush with the Energy E = 1J hits a ideally reflecting mirror, with mass $$m_m = 2 * 10^(-5) kg$$. The light has a wave length of 696 nm. The mirror is hanging on a rope with the length l = 0,1m.

Homework Equations

How big is the deflection of the mirror.

The Attempt at a Solution

I think that tan $$\alpha \approx F_G/F_L$$, where F_G is the Gravitation, F_L the Force caused by the Photons. So I try to get F_L. According to the law of conservation of momentum:
$$I_s = 2 p_l;$$
I_s is the impulse on the mirror and p_l is the sum of all photons' momentum. I take it twice, one time when they arrive at the mirror, a second time when they are reflected.
$$I_s = h/(\lambda) = F * dt$$ so
$$F = h/(\lambda * dt)$$.
I do not know how to get a time, to get the force acting on the mirror

Maybe my approach is completely false, please tell me then.
Thanks and Greetz,
caspar

p.s. excuse my English, I am not a native speaker

 

[Dick]

You can't get a force or a dt since you weren't given the details of the shape of the laser pulse. You only know the combined F*dt which is the impulse (change in momentum - same as the momentum change of the beam). The picture you should have is not of a mirror statically balanced by the force of the beam but one where the mirror gets a one time kick from the beam and starts swinging. The amplitude of the oscillation is the deflection. Does that help?

 

[caspar]

Actually, my problem is to combine the Impulse with the amplitude of the oscillation. However, I've tried another approach:
I assume that the energy the mirror gets is E = mc^2. I calculate the number of emitted photons, get m and receive E. Now, If I translate this Energy E into potential Energy of the mirror, it lifts 5,096*10^3m! I'm afraid that's wrong. I need another approach.

 

[Dick]

But the energy the mirror 'gets' can't be over 1J?? Look I=2*p right? Where p is momentum of the light beam. Doesn't that mean that you know the momentum of the mirror after the light hits it? Doesn't that in turn mean you know the velocity of the mirror immediately after the light hits it.

 

[caspar]

Doesn't that in turn mean you know the velocity of the mirror immediately after the light hits it.

Ohh, so i can get the kinetic energy:
$$v_m = (2*h)/(\lambda*m_m)$$,
which equals the potential energy at the maximum deflection. So the height of the vertical displacement is $$h = (v_m^2*m_m)/(m_m*g)$$. Is that right?

 

[Dick]

Your picture of the problem is now exactly right. But some of those formula look like you just stuck stuff together to get the dimensions right. Eg. your expression for v is independent of the beam energy E (?!). Tell me how to compute the momentum in the beam.

 

[Dick]

Ok. I see what you've done. It's looks correct but is only the displacement for a single photon striking the mirror.

 

[caspar]

the number of photons equals $$n = 1J/(f*h)$$ so
$$v_m = (2*n*h)/(\lambda*m_m)$$
Well, the deflection measured on the level of the unmoved mirror is about s = 3,3667*10^-5m. I feel quite uncomfortable with that number. Is it likely to be such a number?

 

[Dick]

Quite likely. There is not much momentum in 1J of light as you now know.

 

[caspar]

Thank you, Dick, you've helped me a lot.
caspar

 

[Dick]

Your very welcome. It was fun.