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Question

To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is MgF2 (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum (5500oA) there is maximum transmission.

A
5000oA
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B
2000oA
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C
1000oA
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D
3000oA
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Solution

The correct option is C 1000oA
Consider a ray incident at angle i. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r1. Of course, successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence rays r1 and r2 shall dominate the behavior. If incident light is to be
transmitted through the lens, r1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence there is no phase change on reflection. The optical path difference between r2 and r1 is n(AD +CD) - AB.
If d is the thickness of the film , then
AD=CD=dcosr
AB=ACsini
AC2=dtanr
AC=2dtanr

Hence, AB=2dtanr×sini
Thus the optical path difference is
2ndcosr2dtanr×sini

= 2sinisinrdcosr2dsinrcosrsini

=2dsini[1sin2rsinrcosr]=2ndcosr

For these waves to interfere destructively this must be λ \2.
2ndcosr=λ2 ot ndcosr=λ4
For a camera lens, the sources are in the vertical plane and hence
ir0
ndλ4
d=55000A1.38×410000A


1030367_944160_ans_56018eb82ff04ebcb7775f966dc9a15c.png

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