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

$E=\frac{hC}{\lambda }P=\frac{h}{\lambda } \Rightarrow P=\frac{E}{C}$
Initial momentum of light when falls on a perfectly reflecting surface.
$P_i=\frac{E}{C}$
When it is reflected momentum is in opposite direction
$P_f=-\frac{E}{C}$
Net momentum $=P_f-P_i$
$= -\frac{E}{C}-\frac{E}{C}$
$=-\frac{2E}{C}$
$\mathrm{\Delta }P=\frac{2E}{C}$