A plane light wave of intensity `I=0.20Wcm^-2` falls on a plane mirror surface with reflection coefficient `rho=0.8`. The angle of incidence is `45^circ`. In terms of corpuscular theory, find the magnitude of the normal pressure exerted on that surface.

Options

A. `1.2N cm^-2`
B. `0.2Ncm^-2`
C. `2.6Ncm^-2`
D. `0.5N cm^-2

Answer

D

Explanation

Momentum corresponding to incident protons normal to the surface.
`((dP)/(dt))_("incident")=(I)/(c )dAcos^2theta`
Since reflection coefficient is `rho`, so the momentum of the reflected photons per second normal to surface,
`((dP)/(dt))_("reflected")=-(I)/(c )dAcos^2theta`
Hence, rate of change of momentum of the photons,
`((dP)/(dt))_("photons")=-(I)/(c )dA(rho+1)cos^2theta`
From Newton's third law,
`((dP)/(dt))_("surface")=-(I)/(c )dA(rho+1)cos^2theta`
Hence, pressure exerted on surface,
`P=(dF)/(dA)=(I)/(c )dA(rho+1)cos^2theta`