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To ensure almost 100% transmittivity, ph...

To ensure almost `100%` transmittivity, photographic lenses are often coated with a thin layer of dielectric material, like `MgF_(2)(mu=1.38)` . The minimum thickness of the film to be used so that at the centre of visible spectrum `(lambda = 5500 Å)` there is maximum transmission.

A

5000 Ã…

B

2000 Ã…

C

1000 Ã…

D

3000 Ã…

Text Solution

Verified by Experts

The correct Answer is:
C
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 `r_(2)` parallel to `r_(1)`. Of course successive reflections and transmissions will keep on decreasing the amplitude fo the wave. Hence rays `r_(1)` and `r_(2)` shall dominate the behavior. If incident light is to be transmitted through the lens, `r_(1)` and `r_(2)` 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 `r_(2)` and `r_(1)` is `n(AD+CD)-AB.`
If d is the thickness of the film, then

`AD=CD=(d)/("cos"r)`
`AB=AC"sin"i`
`(AC)/(2)=d"tan"r`
` :. AC=2d"tan"r`
Hence, `AB=2d "tan"r"sin"i`
Thus the optical path difference is
`2n(d)/("cos"r)-2d" tan"r"sin"i`
`=2.("sin"i)/("sin"r)(d)/("cos"r)-2d("sin"r)/("cos"r)"sin"i`
` = 2d" sin"i[(1-"sin"^(2)r)/("sin"r"cos"r)]=2nd" cos"r`
For these waves to interfere destructively this must be `lambda//2`.
`2nd" cos"r=(lambda)/(2) " or " nd" cos"r=(lambda)/(4)`
For a camera lens, the sources are in the vertical plane and hence
`i~=r~=0`
` :. nd~=(lambda)/(4)`
`impliesd=(5500Ã…)/(1.38xx4)~=1000Ã…`
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