Question #116105
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
1
Expert's answer
2020-05-18T10:06:29-0400

In our case for destructive interference


2tn2sin2iλ2=(2k+1)λ22t\sqrt{n^2-\sin^2i}-\frac{\lambda}{2}=(2k+1)\frac{\lambda}{2} \to


2tn2sin2i=kλ+λ2+λ2=(k+1)λ,k=0,1,2,....2t\sqrt{n^2-\sin^2i}=k\lambda+\frac{\lambda}{2}+\frac{\lambda}{2}=(k+1)\lambda,k=0,1,2,....


Assume that k=0k=0


2tn2sin2i=λt=λ2n2sin2i=2t\sqrt{n^2-\sin^2i}=\lambda\to t=\frac{\lambda}{2\sqrt{n^2-\sin^2i}}=


=55010921.252sin230°=2.4107m=\frac{550\cdot10^{-9}}{2\sqrt{1.25^2-\sin^230°}}=2.4\cdot10^{-7}m





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