Question

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to two surface (c-velocity of light) is _____

• A. $\frac{2E}{c}$
• B. $\frac{2E}{c^2}$
• C. $\frac{E}{c^2}$
• D. $\frac{E}{c}$

Answer 1

The correct option is A $\frac{2E}{c}$

as moment of a electromagnetic

radiation of energy $E$ is given by

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

c- speed of light

$\Rightarrow$ incident momentum

on the surface $P_i=\frac{E}{C}$

as surface is perfectly reflecting so total moment reflected back

= total momentum after reflection

$P_f=\frac{E}{C}$

momentum transferred to surface

=change in momentum

$=|P_f-P_i|$

as if is in opposite direction

$=P_f=\frac{-E}{C}$

momentum transferred to surface

$=|-\frac{E}{C}-\frac{E}{C}|=\left|\frac{-2E}{C}\right|$

$=\frac{2E}{C}$