Question

Magnesium fluoride $\left(Mg{F}_2\right)$ is generally used as anti-reflection coating. If refractive index of $Mg{F}_2$ is $1.25$, then minimum thickness of film required is (Take $\lambda =500nm$)

• A. $125nm$
• B. $75nm$
• C. $100nm$
• D. $225nm$

Answer 1

The correct option is C $100nm$

Some part of the light gets reflected from the coating surface as light 1 and the other part gets refracted which further gets reflected from glass and after refraction ,emerges as light 2 and interfere with light 1.

For an anti reflection, optical path diffrence $=(m+\frac{1}{2})\lambda$

Let $d$ be the thickness of the coating.

$2n_{1}dcos\theta =(m+\frac{1}{2})\lambda$

For minimum thickness put $m=0$ and for normal incidence $\theta =0^o$

Given $\lambda =500nm$ and $n_1=1.25$

$d=100nm$