Question

A parallel beam of monochromatic light of wavelength $663nm$ is incident on a totally reflecting plane mirror. The angle of incident is ${60^ \circ }$ and the number of photons striking the mirror per second is $1.0 \times {10^{19}}$. If the force exerted by a light beam on the mirror is $Y \times {10^{ - 8}}N$ . Find $Y$ .

Answer 1

We can solve this by Newton's second law, which states that force is equal to rate of change of momentum. And we know that force is the rate of change of momentum. First, we will find the momentum and then put the values in force which is given by rate of change of momentum and then we will get the required solution.

Formula used:
$P = \dfrac{h}{\lambda }$
$P$ is the momentum,
$h$ is the Planck’s constant and
$\lambda$ is the DE Broglie constant.

Complete answer:
According to the question,
Wavelength of monochromatic light, $\lambda = 663 \times {10^{ - 9}}m$
Angle of incidence, $\theta = {60^ \circ }$
Number of photons per second, $n = 1.0 \times {10^{19}}$
Since light ray is incident at the angle ${60^ \circ }$ so there are two components of the momentum
Initial momentum of horizontal direction is $P\sin \theta$
Initial momentum of vertical direction is $P\cos \theta$
Final momentum of horizontal direction is $P\sin \theta$
Final momentum of vertical direction is $- P\cos \theta$
We include negative sign because direction of the final momentum is opposite
So, change in momentum in horizontal direction is $P\sin \theta - P\sin \theta = 0$
And change in momentum in vertical direction is $P\cos \theta - ( - P\sin \theta ) = 2P\cos \theta$
And we know that Momentum of photon is given by,
$P = \dfrac{h}{\lambda }$
$p = \dfrac{{6.63 \times {{10}^{ - 34}}}}{{663 \times {{10}^9}}} = {10^{ - 27}}$
So, force exerted on the wall is

$${\text{force = }}\dfrac{{{\text{Change in momentum}}}}{{{\text{Time taken}}}} \\\ {\text{force = Number of photons per second}} \times {\text{Change in momentum}} \\\$$

So,
$F = n \times [P\cos \theta - ( - P\cos \theta )] \\\ \Rightarrow F = 2nP\cos \theta \\\ \Rightarrow F = 2 \times 1 \times {10^{19}} \times {10^{ - 27}} \times \dfrac{1}{2} \\\ \therefore F = 1 \times {10^{ - 8}}N \\\$
Hence, the force exerted by the light beam on the mirror is $1 \times {10^{ - 8}}N$ .

Note:
The horizontal component of the initial and final momentum (of the photons) is the same. The only part changing is the vertical component. And the relationship between momentum and wavelength for matter waves is given by $P = \dfrac{h}{\lambda }$ .