The optical properties of a medium are governed by the relative permittivity $\left(\varepsilon_r\right)$ and relative permeability $\left(\alpha_r\right)$. The refractive index is defined as $\sqrt{\alpha_r \varepsilon_r}=n$. For ordinary material, $\varepsilon_r>0$ and $\alpha_r>0$ and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with $\varepsilon_r<0$ and $\alpha_r<0$. Since, then such metamaterials have been produced in the laboratories and their optical properties studied. For such materials $n=-\sqrt{\alpha_r \varepsilon_r}$. As light enters a medium of such refractive index the phases travel away from the direction of propagation.
(i) According to the description above show that if rays of light enter such a medium from air (refractive index $=1$ ) at an angle $\theta$ in 2 nd quadrant, then the refracted beam is in the 3rd quadrant.
(ii) Prove that Snell's law holds for such a medium.
Let us assume that the given postulate is true, then two parallel rays would proceed as shown in the figure below
(i) Let $A B$ represent the incident wavefront and $D E$ represent the refracted wavefront. All points on a wavefront must be in same phase and in turn, must have the same optical path length.
$$\begin{aligned} \text{Thus}\quad -\sqrt{\varepsilon_r \alpha_r} A E & =B C-\sqrt{\varepsilon_r \alpha_r} C D \\ \text{or}\quad B C & =\sqrt{\varepsilon_r \alpha_r}(C D-A E) \\ B C & >0, C D>A E \end{aligned}$$
As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2)
Then,
$$\begin{aligned} -\sqrt{\varepsilon_r \alpha_r} A E & =B C-\sqrt{\varepsilon_r \propto_r} C D \\ \text{or}\quad B C & =\sqrt{\varepsilon_r \alpha_r}(C D-A E) \end{aligned}$$
If $B C>0$, then $C D>A E$
which is obvious from Fig (i).
Hence, the postulate reasonable.
However, if the light proceeded in the sense it does for ordinary material, (going from 2nd quadrant to 4th quadrant) as shown in Fig. (i)., then proceeding as above,
$$\begin{aligned} -\sqrt{\varepsilon_r \propto_r} A E & =B C-\sqrt{\varepsilon_r \propto_r} C D \\ \text{or}\quad B C & =\sqrt{\varepsilon_r \propto_r}(C D-A E) \end{aligned}$$
As $A E>C D$, therefore $B C<0$ which is not possible. Hence, the given postulate is correct.
(ii) From Fig. (i)
$$\begin{array}{rlr} B C & =A C \sin \theta_i \\ \text{and}\quad C D-A E & =A C \sin \theta_r \\ \text{As}\quad B C & =\sqrt{\alpha_r \varepsilon_r} \\ \therefore\quad A C \sin \theta_i & =\sqrt{\varepsilon_r \alpha_r} A C \sin \theta_r \\ \text{or}\quad \frac{\sin \theta_i}{\sin \theta_r} & =\sqrt{\varepsilon_r \alpha_r}=n \end{array} \quad[C D-A E=B C]$$
Which proves Snell's law.
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, $AB=2d\tan r\sin i.$
$$\begin{aligned} \text { Thus, the optical path difference } & =\frac{2 n d}{\cos r}-2 d \tan r \sin i \\ & =2 \cdot \frac{\sin i d}{\sin r \cos r}-2 d \frac{\sin r}{\cos r} \sin i \\ & =2 d \sin \left[\frac{1-\sin ^2 r}{\sin r \cos r}\right] \\ & =2 n d \cos r \end{aligned}$$
$$\begin{aligned} & \text { For these waves to interfere destructively path difference }=\frac{\lambda}{2} \text {. } \\ & \Rightarrow \quad 2 n d \cos r=\frac{\lambda}{2} \\ & \Rightarrow \quad n d \cos r=\frac{\lambda}{4}\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\text { For photographic lenses, the sources are normally in vertical plane }\\ &\begin{aligned} & \therefore \quad i=r=0 \Upsilon \\ & \text { From Eq. (i), } \quad n d \cos 0 \Upsilon=\frac{\lambda}{4} \\ & \Rightarrow \quad d=\frac{\lambda}{4 n} \\ & =\frac{5500 \mathop A\limits^o}{4 \times 1.38} \approx 1000 \mathop A\limits^o \end{aligned} \end{aligned}$$