question_answer

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = velocity of light)

A)  \[\frac{E}{C}\]

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 \[=P-{{P}^{1}}=\frac{E}{C}-\left( -\frac{E}{C} \right)=\frac{2E}{C}\]