Question

A monochromatic parallel beam of light of wavelength 6000 Å travelling in air falls on a thin film of refractive index $\sqrt{3}$. The angle of incidence of the beam is 60°. Find the minimum thickness (in nm) of the film such that the reflected light is most intense.

Answer 1

100

Answer 2

The condition for maximum intensity (constructive interference) in reflected light from a thin film, considering a phase change of $\pi$ at the first reflection (air-film interface) and no phase change at the second reflection (film-air interface), is $2\mu t \cos r = \frac{(2m-1)\lambda}{2}$. For minimum thickness, we set $m=1$, giving $2\mu t \cos r = \frac{\lambda}{2}$.

Using Snell's Law, $1 \cdot \sin 60^\circ = \sqrt{3} \sin r$, we find $r=30^\circ$, so $\cos r = \frac{\sqrt{3}}{2}$.

Substituting the given values: $2 \cdot (\sqrt{3}) \cdot t \cdot \left(\frac{\sqrt{3}}{2}\right) = \frac{6000 \, \text{Å}}{2}$.

This simplifies to $3t = 3000 \, \text{Å}$, so $t = 1000 \, \text{Å}$.

Converting to nanometers, $t = 1000 \times 0.1 \, \text{nm} = 100 \, \text{nm}$.