Thank you for reporting, we will resolve it shortly
Q.
A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
Electromagnetic Waves
Solution:
Initial momentum of surface, $p_{i}=\frac{E}{c}$
where, $c=$ velocity of light (constant) and $E$ is energy
Since, the surface is perfectly reflecting, so the same momentum will be reflected completely.
Final momentum, $p_{f}=\frac{-E}{c} $ (negative value)
$\therefore $ Change in momentum, $\Delta p=p_{f}-p_{i}$
$=-\frac{E}{c}-\frac{E}{c}=-\frac{2 E}{c}$
Thus, momentum transferred to the surface is
$\Delta p^{\prime}=|\Delta p|=\frac{2 E}{c} \text {. }$