A) \[\frac{E}{C}\]
B) \[\frac{2E}{C}\]
C) \[\frac{2E}{{{C}^{2}}}\]
D) \[\frac{E}{{{C}^{2}}}\]
Correct Answer: B
Solution :
Momentum of light falling on reflecting surface \[p=\frac{E}{C}\] As surface is perfectly reflecting so momentum reflect \[p'=-\frac{E}{C}\] So momentum transferred \[=P-{{P}^{1}}=\frac{E}{C}-\left( -\frac{E}{C} \right)=\frac{2E}{C}\]You need to login to perform this action.
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