Question

What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 636 nm light? The index of refraction of the coating material is 1.41 and the index of the glass is 1.66.

Answer 1

Concepts and reason

The concept required to solve the given problems is conditions for constructive and destructive interference.

Use the condition for destructive interference to calculate the minimum thickness of coating on lens.

Fundamentals

The condition for destructive interference is given as follows:

$2t = \left( {m + \frac{1}{2}} \right)\frac{\lambda }{n}$

Here, t is thickness, m is order of interference pattern, n is refractive index of medium, and $\lambda$is wavelength.

Rearrange the condition for destructive interference condition $2t = \left( {m + \frac{1}{2}} \right)\frac{\lambda }{n}$ for thickness t of the coating.

$t = \frac{1}{2}\left( {m + \frac{1}{2}} \right)\frac{\lambda }{n}$

Substitute 0 for m and solve for t.

$\begin{array}{c}\\t = \frac{1}{2}\left( {0 + \frac{1}{2}} \right)\frac{\lambda }{n}\\\\ = \frac{\lambda }{{4n}}\\\end{array}$

Substitute 636 nm for $\lambda ,$and 1.41 for n in the above equation $t = \frac{\lambda }{{4n}}$and solve for t.

$\begin{array}{c}\\t = \frac{{636\,{\rm{nm}}}}{{4\left( {1.41} \right)}}\\\\ = 112.7\,{\rm{nm}}\\\end{array}$

Ans:

The minimum thickness of coating is 112.7 nm.