Question

A parallel beam of light is incident normally on a plane surface absorbing 40% of the light and reflecting the rest. If the incident beam carries 60 W of power, the force exerted by it on the surface is:

• A. $3.2\times {10}^{-8}N$
• B. $3.2\times {10}^{-7}N$
• C. $5.12\times {10}^{-7}N$
• D. $5.12\times {10}^{-8}N$

Answer 1

Answer: (B)

$3.2\times {10}^{-7}N$

Momentum of a photon is given as,

$\rho =E/c$

Momentum of the incident photons per second is,

${\rho }_i=Power/c=60/3\times {10}^8=20\times {10}^{-8}$

As power of reflected light is $60\%of60W=36W$, Momentum of the reflected photons per second is,

${\rho }_r=36/3\times {10}^8=12\times {10}^{-8}$, in the opposite direction to ${\rho }_i$

Force exerted by it on the surface is equal to change in momentum per second.

Force, $F={\rho }_i-(-{\rho }_r)=32\times {10}^{-8}=3.2\times {10}^{-7}N$

Option B is correct.