In Fig. we have shown a dielectric film of thickness `d` deposited on a glass lens. Refractive index of film `= 1.38` and refractive index of glass `= 1.5`.
Given, `lambda = 5500 Å`.
A ray is incidence on the film at angle `i`. It is reflected partially along `AB (as r_(1))` and refracted partially along `AD` in the film at angle `r`.
The refracted ray is reflected partially from film glass interface along `DC` and partially transmitted through the glass lens. At `C`, the ray is reflected partially in the film and refracted into air as `r_(2)` parallel to `r_(1)`.
As amplitude of wave goes on decreasing during successive reflections, therefore, rays `r_(1) and r_(2)` dominate the behaviour.
For maximum transmission through the centre of the lens, `r_(1)` and `r_(2)` should interfere destructively.
As both the reflections at `A and D` are from lower to higher refractive index, there is no phase change on reflection.
The optical path difference between `r_(2)` and `r_(1) = n(AD + CD) - AB`
In `DeltaADE`, `cos r = (DE)/(AD) = (d)/(AD) :. AD = (d)/(cos r)`
Similarly, `CD = (d)/(cos r)`
Again, `tan r = (AE)/(DE) = ((1)/(2)AC)/(d) :. AC = 2d tan r`
In `Delta ABC`, `sin i = (AB)/(AC) AB = AC = sin i = 2d tan r = sin i`
Putting in (i), we get Optical path difference between `r_(2)` and `r_(1)`
`x = n((2d)/( cos r)) - 2d tan r sin i`
`= (sin i)/(sin r)(2d)/(cos r) - 2d(sin r)/(cos r ) sin i`
`= 2d sin i ((1 - sin^(2) r)/(sin r cos r))`
`x = 2d = (sin i)/(sin r) cos r = 2 dn cos r`
The waves `r_(2)` and `r_(1)` will interfere destructively when their path difference
`x = (lambda)/(2), i.e., 2 dn cos r = (lambda)/(2) dn cos r = (lambda)/(4)` ...(ii)
For photographic lenses, the sources are normally in vertical plane. `:. i = r = 0^(@)`
Form (ii), `dn cos 0^(@) = (lambda)/(4) d = (lambda)/(4 n) = (5500 Å)/(4 xx 1.38) = 1000 Å`
