Question

Eyeglasses are often coated with an anti-reflecting coating to reduce glare in the visible part of the spectrum. Consider a 95.0-nm-thick coating applied to the lens, where we wish to reduce the glare of the blue end of the spectrum with wavelengths near 450 nm.

a) If the coating’s index of refraction is smaller than that of the lens, what index of refraction for the anti-reflection coating is needed for the thinnest film possible? b) If the coating’s index of refraction is greater than that of the lens, what index of

refraction for the anti-reflection coating is desired for an m = 1 destructive interference scenario?

Answer 1

Thickness = 95.0 nm

wavelength = 450 nm

a) n(coating) < n( lens)

path difference = 2 x n x t = ( 2m+1) (lambda/2 )

for smallest thickness and destructive interference put m= 0

2 x n x 0.95 nm = 450 nm

n = 450/(2*95) =2.37

b

for this part , (i)there is additional path difference of lambda/2

(ii) the destructive interference is for m = 1

2 x n x t + lambda/2 = (2m+1) ( lambda/2)

2 x n x t = (2m-1) ( lambda/2)

2 x n x 95 nm = 450 nm /2

n = 450 / ( 4 x 95) = 1.18