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} $
