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

Answer

2E/c

Initial momentum of surface
p_i = E/c
where c = velocity of light (constant)
Since, the surface is perfectly so, the same momentum will be reflected completely
Final momentum
p_f= E/c
therefore
Change in momentum
∆p = p_f − pi
= -E/c - E/c = -2E/c
Thus, momentum transferred to the surface is
∆p' = |∆p| = 2E/c