Question
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is
-
E/c
-
2E/c
-
Ec
-
E/c2
Solution
B.
2E/c
Initial momentum of surface
pi = E/c
where c = velocity of light (constant)
Since, the surface is perfectly so, the same momentum will be reflected completely
Final momentum
pf= E/c
therefore
Change in momentum
∆p = pf − pi
= -E/c - E/c = -2E/c
Thus, momentum transferred to the surface is
∆p' = |∆p| = 2E/c
