Correct Answer - D
Power of light received by the cone `=(piR^2)`
Let number of photons hitting the cone per second is n.
Then, `nE=IpiR^2impliesn=(piR^2I)/(E)`
By symmetry, the net force on the cone will be vertically downward.
Force due to one photon:
`f=2(h)/(lamda)sintheta`
This force is perpendicular to the surface of cone. Hence, net force on the cone will be
`F=nfsintheta=n(2(h)/(lamda)sintheta)sintheta`
`=(piR^2I)/(c)(1-cos2theta)`