How thick should the coating of a material with index of refraction 1.40 on a pi

Question

How thick should the coating of a material with index of refraction 1.40 on a piece of flat transparent material of index of refraction 1.52 if we wish that there is no reflection for light of wavelength 690 nm? Provide your answer in nm. How thick should the coating of a material if we wish that there is maximum transmission for light of wavelength 690 nm? Provide your answer in nm What if we want to place the coating over a transparent material of index of refraction 1.16 nm and we wish that there is maximum 690 nm, what should the thickness of the coating be? Provide your answer in nm.

Explanation / Answer

Part A)

Apply 2nt = (m + .5)(wavelength)

2(1.4)(t) = (.5)(690 X 10-9)

t = 123 nm

Part B)

Maximum transmission is the same thing as no reflection, Its the same question as part A, asked in a different way to see if you understand.

Thus the answer is still 123 nm

Part C)

In this case the formula changes to

2nt = m(wavelength)

2(1.4)(t) = (1)(690 X 10-9)

t = 246 nm

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