A) \[\frac{2E}{Atc}\]
B) \[\frac{E}{2c}\]
C) \[\frac{2E}{ct}\]
D) zero
Correct Answer: C
Solution :
[c] Incident momentum, \[p=\frac{E}{c}\] For perfectly reflecting surface with normal incidence \[\Delta p=2p=\frac{2E}{c}\]; \[F=\frac{\Delta p}{\Delta t}=\frac{2E}{ct}\];
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