To ensure almost $100 \%$ transmittivity, 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 $\mathrm{MgF}_2(n=1.38)$. What should the thickness of the film be so that at the centre of the visible spectrum ( $5500 \mathop A\limits^o$ ) there is maximum transmission.
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,
$$\lambda=5500 \mathop A\limits^o .$$
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 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 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 $r_2$ 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
$$n(A D+C D)-A B$$
If $d$ is the thickness of the film, then
$$\begin{aligned} A D & =C D=\frac{d}{\cos r} \\ A B & =A C \sin i \\ \frac{A C}{2} & =d \tan r \\ \therefore\quad A C & =2 d \tan r \end{aligned}$$
Hence, $AB=2d\tan r\sin i.$
$$\begin{aligned} \text { Thus, the optical path difference } & =\frac{2 n d}{\cos r}-2 d \tan r \sin i \\ & =2 \cdot \frac{\sin i d}{\sin r \cos r}-2 d \frac{\sin r}{\cos r} \sin i \\ & =2 d \sin \left[\frac{1-\sin ^2 r}{\sin r \cos r}\right] \\ & =2 n d \cos r \end{aligned}$$
$$\begin{aligned} & \text { For these waves to interfere destructively path difference }=\frac{\lambda}{2} \text {. } \\ & \Rightarrow \quad 2 n d \cos r=\frac{\lambda}{2} \\ & \Rightarrow \quad n d \cos r=\frac{\lambda}{4}\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\text { For photographic lenses, the sources are normally in vertical plane }\\ &\begin{aligned} & \therefore \quad i=r=0 \Upsilon \\ & \text { From Eq. (i), } \quad n d \cos 0 \Upsilon=\frac{\lambda}{4} \\ & \Rightarrow \quad d=\frac{\lambda}{4 n} \\ & =\frac{5500 \mathop A\limits^o}{4 \times 1.38} \approx 1000 \mathop A\limits^o \end{aligned} \end{aligned}$$