A radiation of energy `E` falls normally on a perfctly refelecting surface . The momentum transferred to the surface is

Options

A. `E//c`
B. `2E//c`
C. `Ec`
D. `E//c^(2)`

Answer

B

Explanation

Initial momentum of surface
`P_(i) = (E)/(C)`
where c = velocity of light (constant). Since, the surface is perfectly reflecting so the same momentum will be reflected completely Final momentum
`P_(i)=(E)/(C)`(negative value)
`:.` Change in momentum
`Delta_(p) = p_(f)-P_(i) = -(E)/(C)-(E)/(C) = -(2E)/(C)`
Thus, momentum transferred to the surface is
`Delta_(p)=|Delta_(p)|=(2E)/(C)`.