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.

Explanation

In this figure, we have shown a dielectric film of thickness d deposited on a glass lens.

Refreactive index of film= 1.38 and refractive index of glass = 1.5.
Given , `lambda=5500Å`.
Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface and a part refracted inside.
This is a 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 tansmitted as `r_(2)` parallel to `r_(1)`. Of course successive reflection and transmission will keep on decreasing the amplitude of the wave.
Hence, rays `r_(1)` and `r_(2)` shall dominate the behaviour. If incident light is to be transmitted through the lens `r_(1)` and `r_(2)` shuld interfere destructively. Both the reflection at A and D are from lower to highet 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 `=(2nd)/( cos r)-2d tan r sin i `
`= 2 . (sin i d)/(sin r cos r)- 2d (sin r)/(cos r) sin i`
`=2d sin [(1- sin^(2) r)/(sin r cos r)]`
`=2nd cos r`
For these waves to interfere destructively path difference `=(lambda)/(2)`
`implies 2nd cos r=(lambda)/(2)`
`implies nd cos r =(lambda)/(4)`.....(i)
For photographic lenses, the sources are normally in vertical plane.
`:. i=r=0^(@)`
From Eq.(i) `nd cos 0^(@)=(lambda)/(4)`
`implies d=(lambda)/(4n)`
`=(5500Å)/(4xx1.38)~~1000Å`