Question

A small mirror of area 'A' and mass 'm' is suspended in a vertical plane by means of a weightless string. A beam of light of intensity 'I' falls normally on the mirror and the string is deflected form the vertical through a very small angle ' $\alpha$ '. Assuming the mirror to be perfectly reflecting, obtain an expression for $\alpha$

• A. $\frac{2IA}{cmg}$
• B. $\frac{I}{cmg}$
• C. $\frac{3I}{cmg}$
• D. None of the above

Answer 1

Answer: (A)

When the light is incident on the mirror is totally reflected and there is no loss of energy.
$\therefore$ Change in momentum $=2\times$ incident momentum
Intensity $=\frac{\text{ Energy }}{\text{ Time }\times \text{ Area }}$
Momentum $=\frac{\text{ Intensity }}{\text{ Velocity of light }}$
Pressure, $P=\frac{\text{ Force }}{\text{ Area }}$
$=\frac{\text{ Rate of change in momentum }}{A}$
$=\frac{2IA}{c}$( where $c=$ speed of light)
$\alpha =\tan \alpha =\frac{P\times A}{mg}=\frac{2IA}{cmg}$