A light of wavelength of 525 nm is incident normally on a glass coated w/A thin

Question

A light of wavelength of 525 nm is incident normally on a glass coated w/A thin coating of anti-Respective material. If the glass has an index of refraction of 1.76 phi. The coating has an index of refraction of 1.26. Find a) the wavelength of light in the glass b) in the coating c) What is the minimum thickness of the coating that will produce a destructive interference of the reflective light rays from the two surfaces of coating d) What is the minimum thickness of the coating that will produce a constructive interference of the reflected light rays from the two surfaces of the coating.

Explanation / Answer

wavelength of light in glass = wavelength of light in air/refractive indexof glass

wavelength of light in glass = 525/1.76 = 298.3 nm

wavelength of light in coating = 525/1.26 = 416.7 nm

(c)

In the coating

path difference = 2*t

lambda in glass = lambda in air/nglass

for destructive interference path diference = m*lambda in glass

2t = m*lambda in air/nglass

t = m*lambda in air/(2n)

for m = 1

t = 1*525*10^-9/(2*1.76)

t = 149 nm   <<<<----answer

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d)

for constructive interference path diference = m*lambda in glass/2

2t = m*lambda in air/(2*nglass)

t = m*lambda in air/(4n)

for m = 1

t = 1*525*10^-9/(4*1.76)

t = 74.5 nm   <<<<----answer

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