A material having an index of refraction of 1.25 is used as an antireflective co

Question

A material having an index of refraction of 1.25 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 580 nm light?

116

If instead a material with an index of refraction of 2.05 is used for the coating, what are the two smallest thicknesses to minimize reflection (in order).

0

141

A material having an index of refraction of 1.25 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 580 nm light? If instead a material with an index of refraction of 2.05 is used for the coating, what are the two smallest thicknesses to minimize reflection (in order).

Explanation / Answer

Part A)

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

2(1.25)(t) = (.5)(580 X 10^-9)

t = 1.16 X 10^-7m which is 116 nm

Part B)
The equation switches to 2nt = m(wavelength)

2(2.05)(t) = 1(580 X 10^-9)

t = 1.414 X 10^-7 m

141.4 nm

Also

2(2.05)(t) = 2(580 X 10^-9)

t = 2.83 X 10^-7 m

That is 283 nm

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