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

$\frac{2E}{c}$

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

$Δp' = |Δp| =\frac{2E}{c}$