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Question: A material having an index of refraction of 1.20 is used as an antireflective coating on a piece ...

A material having an index of refraction of 1.20 is used as an antireflective coating on a piece of glass (n = 1.50). What should be the minimum thickness of this film in order to minimize reflection of 400 nm light?

1) 83.3 nm (correct answer)

If instead a material with an index of refraction of 1.80 is used for the coating, what are the two smallest thicknesses to minimize reflection (in order).

2 ) __??_____ nm and 3) 111.1 nm (correct answer)

Help, I got the answer to 1 and 3 of this question, but I cannot get part 2 with the ??.

I have tried 55.5, 111.1, 222.2 and 100 which are all incorrect.

I tried following what others had done and everytime part 2 was wrong.

Please help. Thanks

Explanation / Answer

Part 1)

Equation is :-

=> 2 * n * t = (m + 0.5) * (Wavelength)

=> 2 * 1.2 * t = 0.5 * (400*10^-9)

=> t = (2*10^-7) / (2.4)

=> t = 83.33*10^-9

Part 2) Equation switches to 2*n*t = m * (Wavelangth)

2 * (1.80) * (t) = 1 (400*10^-9)

t = 400*10^-9 / 3.6

t = 111.11*10^-9

t = 111.11 nm

Also

2(1.80)(t) = 2(400*10^-9)

t = 800*10^-9 / 3.6

t = 222.22 nm

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