Ans:(a), (c) and (d)

Explanation:

It is defined as the momentum imported per second per unit are on which light falls.

$$\begin{gathered} \mathrm{i}.\mathrm{e}. \boxed{\mathrm{Radiation} \mathrm{pressure} = \frac{\mathrm{Momentum}}{\left(\mathrm{Area}\right) \left(\mathrm{Time}\right)}} \\ \mathrm{But} \mathrm{momentum} \mathrm{transfered} \mathrm{in} \mathrm{EMW} \mathrm{is} \mathrm{P}= \frac{\mathrm{U}}{\mathrm{C}} \\ \mathrm{Force} \mathrm{F}=\frac{\mathrm{P}}{\mathrm{t}}=\frac{{\displaystyle \raisebox{1ex}{$\mathrm{U}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{C}$}\right.}}{\mathrm{t}} = \frac{\mathrm{U}}{\mathrm{C}.\mathrm{t}} \\ \therefore \mathrm{Pressure} \mathrm{P}=\frac{\mathrm{Force}}{\mathrm{Area}}=\frac{\mathrm{F}}{\mathrm{A}} \\ \mathrm{P}=\frac{\frac{\mathrm{U}}{\mathrm{C}.\mathrm{t}}}{\mathrm{A}}=\frac{\mathrm{U}}{\mathrm{A}.\mathrm{C}.\mathrm{t}}=\left(\frac{\mathrm{U}}{\mathrm{At}}\right)\left(\frac{1}{\mathrm{C}}\right) \\ \boxed{ \mathrm{P}=\frac{\mathrm{I}}{\mathrm{C}}} \left[\because \mathrm{I}\frac{\mathrm{U}}{\mathrm{A}.\mathrm{t}}\right] \end{gathered}$$

It is valid for total absorption. 

For totally reflecting surface $\boxed{ P=\frac{2I}{C}}$

Radiation pressure P comes in the range between $\frac{\mathrm{I}}{\mathrm{C}}<p<\frac{2\mathrm{I}}{\mathrm{C}}$