A small mirror of mass m is suspended by a massless thread of length $$l$$. Then the small angle through which the thread will be deflected when a short pulse of laser of energy E falls normal on the mirror (c = speed of light in vacuum and g = acceleration due to gravity)

Solution :

Momentum carried by light pulse of energy E is :

$$p = \frac{E}{c}$$

Since the mirror reflects the light normally, change in momentum of light is :

$$\Delta p = 2\frac{E}{c}$$

Hence momentum imparted to mirror :

$$mv = \frac{2E}{c}$$

Therefore,

$$v = \frac{2E}{mc}$$

After receiving impulse, mirror starts oscillating.

At maximum deflection, kinetic energy converts into potential energy.

So,

$$\frac{1}{2}mv^2 = mgh$$

For small angle $$\theta\ $$,

$$h \approx \frac{l\theta^2}{2}$$

Therefore,

$$\frac{1}{2}mv^2 = mg\frac{l\theta^2}{2}$$

$$v^2 = gl\theta^2$$

Substituting,

$$\left(\frac{2E}{mc}\right)^2 = gl\theta^2$$

$$\theta = \frac{2E}{mc\sqrt{gl}}$$

Final Answer :

$$\theta = \frac{2E}{mc\sqrt{gl}}$$