A radiation of energy E is normally incident on a perfectly reflecting surface. The momentum transferred to the surface will be: (Given: speed of light 'c')
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $2Ec^{-1}$
Energy of radiation = $E$
Momentum of radiation = $\frac{E}{c}$
Since the surface is perfectly reflecting, change in momentum = $2 \times \frac{E}{c}$
Momentum transferred to the surface = $\frac{2E}{c}$
Answer: $\frac{2E}{c}$