Answer to Question #116105 in Optics for Jason

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 $k=0$

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

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