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) Ec (D) (E/c2)

Tardigrade Admin · Accepted Answer

Initial momentum of surface Pi=(E/C) Where, c = velocity of light (constant). 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'=|.Δ p|.=(2 E/c)

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​