A parallel beam of monochromatic light of wavelength 500nm 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

(a) The energy of each photon is `E = hc/ lambda = ((4.14 xx (10^-15)eVs) xx (3 xx (10^8) m (s^-1))/500nm) `
` 1242 eVnM/ 500nm = 2.48 eV.`
In one second, 10J of energy passes through any cross
section of the beam. Thus, the number of photons
crossing a cross section is
` n= (10J/2.48 eV)= 2.52 xx (10^19)`.
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/lambda)= hv/c. `
The total momentum of all the photons falling per
second on the surface is
` = nhv/c = 10J/c = (10J/ 3 xx (10^8)m(s^-1))= 3.33 xx (10^-8) Ns.`
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= (3.33 xx (10^-8)Ns / 1s) = 3.33 xx (10^-8)N`