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')

$2Ec^{-1}$

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: $\frac{2E}{c}$