A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
Answer & explanation
Correct answer: option 2
Initial momentum of surface
$P_i=\frac{E}{C}$
Where, c = velocity of light (constant).
Since, the surface is perfectly, reflecting, so the same momentum will be reflected completely.
Final momentum
$P_f=\frac{E}{C}$ (negative value)
∴ Change in momentum
$Δp = p_f - p_i $
$ = - \frac{E}{c} - \frac{E}{c} = - \frac{2E}{c}$
Thus , momentum transferred to the surface is
$Δp' = |Δp| =\frac{2E}{c}$