A source of light of power P is shown in Fig. 3.6. Find the force on the block placed in the path of the light rays. The surface of body on which the light beam is incident is having a reflection coefficient `a_r=0.7` and absorption coefficient `a_a=0.3`.

Explanation

In this case `70%` of the incident photons are reflected back and `30%` are absorbed by the body. Thus, the photon which is absorbed will impart a momentum `(h)/(lamda)` to the body and the photon which is reflected will impart the change in momentum `(2h)/(lamda)` to the body. Thus, the net force acting on body can be given as
`F=(0.7Plamda)/(hc)xx(2h)/(lamda)+(0.3Plamda)/(hc)xx(h)/(lamda)`
`F=(1.7P)/(c)`