Question

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

• A. $\frac{2E}{C}$
• B. $\frac{2E}{C^2}$
• C. $\frac{E}{C^2}$
• D. $\frac{E}{C}$

Answer 1

Answer: (A)

$\frac{2E}{C}$

The radiation of energy is given by,

$E=\frac{h}{\lambda }$

Initial momentum of the radiation is

$P_i=-\frac{h}{\lambda }=-\frac{E}{C}$

The reflected momentum is

$P_r=-\frac{h}{\lambda }=-\frac{E}{C}$

So, the change in momentum of light is

$\Delta P_{light}=P_r-P_i=-\frac{2E}{C}$

Thus, the momentum transferred to the surface is,

$\Delta P_{light}=P_r-P_i=-\frac{2E}{C}$