Question

A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

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

Answer 1

Answer: (B)

Initial momentum of surface
$ P_{i}=\frac{E}{C}$
Where, $c \, $ = velocity of light (constant).
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
$\Delta p'=\left|\right.\Delta p\left|\right.=\frac{2 E}{c}$