Question

2.) An antireflection coating (n= 1.39) sprayed over a glass lens (n=1.53) is designed to give no reflection at wavelength of 550nm.

b) What is the minimum thickness of the coating?

c) How would the interference change if the index of refraction of the lens is n = 1.35?

d) Find the minimum non-zero thickness of the coating if the index of refraction of the lens is n = 1.35

Answer 1

Solution:

Given data

The refractive index of antireflection coating, $n_{c}=1.39$

The refractive index of glass lens, $n_{g}=1.53$

It is given that,

air 550 nm

The minimum thickness of the film can be calculated as follows

$t=\frac{\lambda _{film}}{4}$

where

$\lambda _{film}=\mathrm{wavelength\; of \; light\; in\; the\; film }$

This can be calculated as follows

$\lambda _{film}=\frac{\lambda _{air}}{n_{c}}$

$\lambda _{film}=\frac{550}{1.39}$

$\lambda _{film}=395.68\; \mathrm{nm}$

Now thickness can be calculated as follows

$t=\frac{\lambda _{film}}{4}$

$t=\frac{395.68}{4}$

$t=98.92\; \mathrm{nm}$

(c)

The concept behind the antireflecting coating is the double reflection within the film, the two reflected waves interfere destructively only when the thickness of the films is the quarter wavelength and its refractive should be lower than the glass lens.

But, when we take the refractive index of the glass lens as $n_{g}=1.35$

Then we have

Therefore, the two reflected waves will not be in 180o out of phase and will not interfere destructively. Due to this, the coating can not block the light of the given wavelength.

(P)

Now take $n_{g}=1.35$

The minimum non-zero thickness of the film can be calculated as follows

$t=\frac{\lambda _{film}}{4}$

where

$\lambda _{film}=\mathrm{wavelength\; of \; light\; in\; the\; film }$

This can be calculated as follows

$\lambda _{film}=\frac{\lambda _{air}}{n_{c}}$

$\lambda _{film}=\frac{550}{1.39}$

$\lambda _{film}=395.68\; \mathrm{nm}$

Now thickness can be calculated as follows

$t=\frac{\lambda _{film}}{4}$

$t=\frac{395.68}{4}$

$t=98.92\; \mathrm{nm}$

Here, we got the same thickness as of part (b), because the thickness of the coating doesn't depend on the refractive index of the glass lens.

This value play a role only whether the two reflected waves will be destructively interfere with or not.