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The non reflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 636-nm light. If the lens glass has an index of refraction of 1.56, what is the minimum thickness of the coating that will accomplish this task

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Answer:

The minimum thickness is
t = 1.0192 *10^(-7) \ m

Step-by-step explanation:

From the question we are told that

The refractive index is
n = 1.29

The wavelength of the light is
\lambda = 636 \ nm = 636 *10^(-9) \ m

The refractive index of the glass lens is
n_g = 1.56

Generally the condition for destructive interference of a films is


2t = [ m + (1)/(2) ] (\lambda)/(n)

for minimum m = 0


2t = [ 0 + (1)/(2) ] (\lambda)/(n)

=>
2t = [ 0 + (1)/(2) ] (636 *10^(-9))/(1.56)

=>
t = 1.0192 *10^(-7) \ m

User Levi Ramsey
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