Question

A small ideal mirror of mass m is suspended by a weightless thread of length l. Find the angle through which the thread will be deflected when a short laser pulse with energy E is shot in the horizontal direction at angles to the mirror.

Answer 1

Question #81222, Physics / Other

A small ideal mirror of mass $m$ is suspended by a weightless thread of length $l$. Find the angle through which the thread will be deflected when a short laser pulse with energy $E$ is shot in the horizontal direction at angles to the mirror.

Solution

$$| \boldsymbol{p}_f - \boldsymbol{p}_i | = \frac{2E}{c}$$

Assume $\boldsymbol{p}_i = \mathbf{0}$. Thus,

$$| \boldsymbol{p}_f | = \frac{2E}{c}$$ $$\frac{p_f^2}{2m} = \frac{2E^2}{mc^2}$$

From the conservation of energy:

$$\frac{2E^2}{mc^2} = mgl(1 - \cos \theta)$$

So,

$$\frac{2E^2}{mc^2} = 2mgl\sin^2\frac{\theta}{2}$$

Or

$$\sin \frac{\theta}{2} = \left(\frac{E}{mc}\right) \frac{1}{\sqrt{gl}}$$

The angle through which the thread will be deflected when a short laser pulse with energy $E$ is shot in the horizontal direction at angles to the mirror:

$$\theta = 2 \sin^{-1} \left[ \left(\frac{E}{mc}\right) \frac{1}{\sqrt{gl}} \right]$$

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