Question

A thin film of thickness $t$ and index of refraction $1 \cdot 33$ coats a glass with index of refraction $1 \cdot 50$. Find the least thickness that will strongly reflect light with wavelength $600{\text{nm}}$ incident normally.
A) ${\text{225nm}}$
B) $300{\text{nm}}$
C) $400{\text{nm}}$
D) $450{\text{nm}}$

Answer 1

When light is incident on a thin film, the light will get partly reflected and partly refracted. We have to find the least thickness of the thin film such that light of the given wavelength when incident normally on the film suffers strong reflection. Since it is mentioned that the incident light has to be reflected strongly, we invoke the condition for the interference pattern to appear bright when viewed in transmitted light.

Formula used:
-For the thin film to appear bright when viewed in transmitted light, $2\mu t\cos r = n\lambda$ where $\mu$ is the refractive index of the film, $r$ is the angle of refraction, $n$ is the order of the fringe and $\lambda$ is the wavelength of the incident light.

Complete step by step answer.
Step 1: List the parameters involved in the given problem.

The refractive index of the thin film is given to be $\mu = 1 \cdot 33$ .
The refractive index of glass is given to be ${\mu _g} = 1 \cdot 50$ .
The wavelength of the incident light is given to be $\lambda = 600{\text{nm}}$ .
The thickness of the thin film is given to be $t$ .
Also, it is mentioned that the light is incidentally normal to the surface so the angle of refraction will be zero i.e., $r = 0^\circ$ .

Step 2: Express the condition for the strong reflection of the incident light.
For the thin film to appear bright when viewed in transmitted light, the condition to be satisfied is expressed as $2\mu t\cos r = n\lambda$ ---------- (1) where $\mu$ is the refractive index of the given thin-film, $r$ is the angle of refraction, $n$ is the order of the fringe and $\lambda$ is the wavelength of the incident light.
For least thickness $t = {t_{\min }}$, we take $n = 1$ .
Then equation (1) becomes $2\mu {t_{\min }}\cos r = \lambda$ .
$\Rightarrow {t_{\min }} = \dfrac{\lambda }{{2\mu \cos r}}$ ------- (2)
Substituting for $\mu = 1 \cdot 33$, $\lambda = 600{\text{nm}}$ and $r = 0^\circ$ in equation (2) we get, ${t_{\min }} = \dfrac{{600}}{{2 \times 1 \cdot 33 \times \cos 0^\circ }} = 225 \cdot 5{\text{nm}}$
$\therefore$ the least thickness of the film is obtained to be ${t_{\min }} = 225{\text{nm}}$ .

So the correct option is A.

Note: Here, the light is incident normally on the air – thin-film interface, so a part of it will get reflected and the other part will be transmitted through the film till it reaches the thin-film – glass interface where the process of reflection and transmission repeats. So refraction of light does not take place. Hence for normal incidence, the angle of refraction $r$ is taken to be zero. If the interference pattern is viewed in the reflected light, then the condition would be different.