A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

Question

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (A) (E/c) (B) (2 E/c) (C) E c (D) (E/c2)

Tardigrade Admin · Accepted Answer

Initial momentum of surface, pi=(E/c) where, c= velocity of light (constant) and E is energy Since, the surface is perfectly reflecting, so the same momentum will be reflected completely. Final momentum, pf=(-E/c) (negative value) ∴ Change in momentum, Δ p=pf-pi =-(E/c)-(E/c)=-(2 E/c) Thus, momentum transferred to the surface is Δ p prime=|Δ p|=(2 E/c) text .

Q. A radiation of energy $E$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

$\frac{E}{c}$

$\frac{2 E}{c}$

$E c$

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

Q. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

cE​

c2E​