Question

Magnesium fluoride $\left(Mg{F}_2\right)$ is generally used as anti-reflection coating.
If refractive index of $Mg{F}_2$ is $1.25$ and if the minimum thickness of film is not technologically possible to manufacture, then next thickness of film required is (approximately) (Take $\lambda =500nm$)

• A. $300nm$
• B. $12nm$
• C. $750nm$
• D. $550nm$

Answer 1

The correct option is A $300nm$

Some part of the ray gets reflected from the coating surface as ray 1 and the other part gets refracted which further gets reflected from glass and after refraction ,emerges as ray 2 and interfere with ray 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 normal incidence $\theta =0^o$

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

Put $m=1$

$d=300nm$