Calculating Force Exerted by Laser Beam on a Mirror

[Matthaeus_]

Homework Statement

A laser beam ($$\mathrm{Power} = 1\ \mathrm{W})$$ is completely reflected by a mirror perpendicular to the beam. Light is made of photons, and each photon carries an energy $$E = h\nu$$ and a momentum $$P = h/\lambda$$, where $$\nu$$ is the frequency, $$\lambda$$ is the wavelength and $$h$$ is Planck's constant. Find the force with which light pushes the mirror.

Homework Equations

Apart from those already present in the problem statement, I have:
$$\lambda \nu = c$$

$$F = \frac{dp}{dt}$$

The Attempt at a Solution

Each second, the light source emits $$n$$ photons, each one carries an energy $$E = h\nu = hc/\lambda$$, for a total power of $$1\ \mathrm{W}$$. This gives:

$$\displaystyle n = \frac P E = \frac{\lambda}{hc}$$

In one second then, $$n$$ photons hit the mirror and bounce back, which gives:

$$\displaystyle F = \frac{dp}{dt} = n \cdot 2p = 2 \frac{\lambda}{hc}\cdot \frac{h}{\lambda} = \frac 2 c \approx 6.67\cdot 10^{-9}\ \mathrm{N}$$

The result is somewhat intuitively pleasing, can you check it is correct, please?

 

[Doc Al]

Looks good to me.

 

[Matthaeus_]

Whew... Thank you Doc for checking :)

 

[faiyth]

Sorry to dig up such an old question, but after looking at this question, I don't understand were the 2 comes from in the F=n2p. Could someone explain this to me?
Thanks

 

[mgb_phys]

The momentum is twice the incoming photon's because it bounces back
Think of a ball, if you throw it to hit a wall and stop then you need twice as much force for it to hit the wall and come back at the same speed.

 

[faiyth]

That is to say, if it was not a mirror, and the light did not reflect off the surface, the 2 would be a 1 instead?

 

[Doc Al]

Sorry to dig up such an old question, but after looking at this question, I don't understand were the 2 comes from in the F=n2p.

It comes from the fact that the change in momentum is twice the original momentum.

 

[Redbelly98]

That is to say, if it was not a mirror, and the light did not reflect off the surface, the 2 would be a 1 instead?

Yes. If the light is absorbed then it's a 1. As Doc Al said, it's all about change in momentum.