Rankers Physics

Properties of EM Waves: Practice Problem & Solution

Radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = velocity of light): (2015)
$\frac{2E}{C}$
$\frac{2E}{C^2}$
$\frac{E}{C^2}$
$\frac{E}{C}$

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Properties of EM Waves. Understanding the underlying formula is key to arriving at the correct answer below:

The momentum of the incident radiation is $p = \frac{E}{C}$.
Since the surface is perfectly reflecting, the radiation reflects back with momentum $-p$.
The momentum transferred to the surface is the change in momentum: $\Delta p = p - (-p) = 2p = \frac{2E}{C}$.

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