### Neet Solved Paper 2015 Question 35

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

#### Options:

A) $ \frac{E}{c} $

B) $ \frac{2E}{c} $

C) $ \frac{2E}{c^{2}} $

D) $ \frac{E}{c^{2}} $

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

The radiation energy is given by $ E=\frac{hc}{\lambda } $

Initial momentum of the radiation is $ {{\mathbf{P}} _{i}}=\frac{h}{\lambda }=\frac{E}{c} $

The reflected momentum is $ {{\mathbf{P}} _{r}}=-\frac{h}{\lambda }=-\frac{E}{c} $

So, the change in momentum of light is $ \Delta {{\mathbf{P}} _{light}}={{\mathbf{P}} _{r}}-{{\mathbf{P}} _{i}}=-\frac{2E}{c} $

Thus, the momentum transferred to the surface is $ \Delta {{\mathbf{P}} _{light}}=\frac{2E}{c} $