Electromagnetic Waves - Result Question 5
5. A radiation of energy ’ $E$ ’ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is $(C=$ Velocity of light $)$
(a) $\frac{2 E}{C}$
(b) $\frac{2 E}{C^{2}}$
(c) $\frac{E}{C^{2}}$
(d) $\frac{E}{C}$
[2015]
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Answer:
Correct Answer: 5. (a)
Solution:
- (a) Momentum of light falling on reflecting surface $p=\frac{E}{C}$ As surface is perfectly reflecting so momentum reflect $p^{\prime}=-\frac{E}{C}$
So, momentum transferred
$=P-P^{\prime}=\frac{E}{C}-(-\frac{E}{C})=\frac{2 E}{C}$
