Question

To ensure almost 100 per cent 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 center of the visible spectrum (5500 Å) there is maximum transmission.

Answer 1

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 is 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 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.

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 tanr sini`

Thus the optical path difference is

`2n  d/cos r - 2d tan r sin i`

= `2. sin i/sin r d/cos r - 2d sinr/cos r sin i`

= `2d sin [(1 - sin^2r)/(sinr cos r)]`

= `2 nd cos r`

For these waves to interfere destructively this must be `λ/2`.

⇒ `2nd cos r = λ/2`

or `nd cos r = λ/4`

For a camera lens, the sources are in the vertical plane and hence `i ≃ r ≃ 0`

∴ `nd ≃ λ/4`

⇒ `d = (5500 Å)/(1.38 xx 4) ≃ 1000 Å`