Answer to Question #117025 in Optics for victoria

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 $k=0$

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