To ensure almost 100% transmissivity, 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 centre of the visible spectrum (5500 A), there is a maximum transmission.

Hint: Use Snell's law.

Step 1: Find the optical path difference.

In this figure, 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, λ=5500 A

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface and a part is 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 r1and r2 shall dominate the behaviour. If the 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.

lf d is the thickness of the film, then;

                                         AD=CD=dcosrAB=ACsiniAC2=dtanrAC=2dtanr

Hence, AB=2dtanrsini.

 Thus, the optical path difference==2ndcosr2dtanrsini=2sini×dsinr×cosr2dsinrcosrsini=2dsini[1sin2rsinrcosr]=2ndcosr

Step 2: Find the conditions for maxima and minima.

For these waves to interfere destructively path difference =λ2


2nd cos r=λ2
nd cos r =λ4

For photographic lenses, the sources are normally in the vertical plane,

                            i=r=0From Eq. (i), ndcos0=λ4 d=λ4nd=55004×1.381000