# Force exerted by a laser beam

## 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?

## Answers and Replies

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Doc Al
Mentor
Looks good to me.

Whew... Thank you Doc for checking :)

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
Homework Helper
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.

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
Mentor
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
Staff Emeritus