Question

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

Answer 1

In our case for destructive interference

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

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

Assume that $k=0$

$2t\sqrt{n^2-\sin^2i}=\lambda\to t=\frac{\lambda}{2\sqrt{n^2-\sin^2i}}=$

$=\frac{550\cdot10^{-9}}{2\sqrt{1.25^2-\sin^230°}}=2.4\cdot10^{-7}m$