A parellel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross section of the beam is 10 W. Find
(a) the number of photons absorbed per second by the surface and
(b) the force exerted by the light beam on the surface.

Explanation

The energy of each photon is
`E=(hc)/(lamda)=((4.14xx10^(-15)eV-s)xx(3xx10^(8)(m)/(s)))/(500nm)`
`=(1242eV-nm)/(500nm)=2.48eV`
In 1 s, 10 J of energy passes through any cross section of the beam. Thus, the number of photons crossing a cross section per unit time is
`n=(10J)/(2.48eV)=2.52xx10^(19)`
This is also the number of photons falling on the surface per second and being absorbed.
b. The linear momentum of each photon is `p=(h)/(lamda)=(hv)/(c )`
The total momentum of all the photons falling per second on the surface is
`(nhv)/(c)=(10J)/(c)=(10J)/(3xx10^(8)(m)/(s))=3.33xx10^(-8)N-s`
As the photons are completely absorbed by the surface, this much momentum is transferred to the surface per second. The rate of change of the momentum of the surface, i.e., the force on it is
`F=(dp)/(dt)=(3.33xx10^(-8))/(1s)=3.33xx10^(-8)N`