Question

The material to be used for an antireflective coating has index of refraction of 1.25. How thick should the coating be to give the best result for l=550 nm and an angle of incidence of 30 degrees with the normal?

Answer 1

For destructive interference (assume that the refractive index of the surface on which the coating is applied $n_0>1.25$ )

$2t\sqrt{n^2-\sin^2i}=(2k+1)\frac{\lambda}{2},k=0,1,2,...$

So, for $k=0$

$2t\sqrt{n^2-\sin^2i}=\frac{\lambda}{2}\to t=\frac{\lambda}{4\sqrt{n^2-\sin^2i}}=\frac{550\cdot10^{-9}}{4\sqrt{1.25^2-\sin^230°}}=1.2\cdot10^{-7}m$