$$A C=C O=D, S_1 C=S_2 C=d \ll D$$
A small transparent slab containing material of $\alpha=1.5$ is placed along $A S_2$ (figure). What will be the distance from 0 of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?
In case of transparent glass slab of refractive index $\propto$, the path difference will be calculated as $\Delta x=2 d \sin \theta+(\alpha-1) L$.
In case of transparent glass slab of refractive index $\propto$, the path difference $=2 d \sin \theta+(\alpha-1) L$.
For the principal maxima, (path difference is zero) i.e.,
$$2 d \sin \theta_0+(\alpha-1) L=0$$
or $\quad \sin \theta_0=-\frac{L(\alpha-1)}{2 d}=\frac{-L(0.5)}{2 d} \quad[\because L=d / 4]$
or $$\sin \theta_0=\frac{-1}{16}$$
$$\begin{aligned} &\therefore \quad O P=D \tan \theta_0 \approx D \sin \theta_0=\frac{-D}{16}\\ &\text { For the first minima, the path difference is } \pm \frac{\lambda}{2}\\ &\therefore \quad 2 d \sin \theta_1+0.5 L= \pm \frac{\lambda}{2} \end{aligned}$$
$$\begin{aligned} \text { or } \quad \sin \theta_1 & =\frac{ \pm \lambda / 2-0.5 L}{2 d}=\frac{ \pm \lambda / 2-d / 8}{2 d} \\ & =\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16} \end{aligned}$$
[ $\because$ The diffraction occurs if the wavelength of waves is nearly equal to the side width $(d)$ ]
On the positive side $\sin \theta_1^{\prime+}=+\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$
On the negative side $\sin \theta^{\prime \prime}{ }_1^{-}=-\frac{1}{4}-\frac{1}{16}=-\frac{5}{16}$
The first principal maxima on the positive side is at distance
$$D \tan \theta_1^{\prime+}=D \frac{\sin \theta_1^{\prime+}}{\sqrt{1-\sin ^2 \theta_1^{\prime}}}=D \frac{3}{\sqrt{16^2-3^2}}=\frac{3 D}{\sqrt{247}} \text { above point } O$$
The first principal minima on the negative side is at distance
$$D \tan \theta^{\prime \prime}{ }_1=\frac{5 D}{\sqrt{16^2-5^2}}=\frac{5 D}{\sqrt{231}} \text { below point } O.$$
Four identical monochromatic sources $A, B, C, D$ as shown in the (figure) produce waves of the same wavelength $\lambda$ and are coherent. Two receiver $R_1$ and $R_2$ are at great but equal distances from $B$.
(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when $B$ is turned off?
(iii) Which of the two receivers picks up the larger signal when $D$ is turned off?
(iv) Which of the two receivers can distinguish which of the sources $B$ or $D$ has been turned off?
Consider the disturbances at the receiver $R_1$ which is at a distanced from $B$. Let the wave at $R_1$ because of $A$ be $Y_A=a \cos \omega t$. The path difference of the signal from $A$ with that from $B$ is $\lambda / 2$ and hence, the phase difference is $\pi$. Thus, the wave at $R_1$ because of $B$ is $$ y_B=a \cos (\omega t-\pi)=-a \cos \omega t . $$ The path difference of the signal from $C$ with that from $A$ is $\lambda$ and hence the phase difference is $2 \pi$. Thus, the wave at $R_1$ because of $C$ is $Y_C=a \cos (\omega t-2 \pi)=a \cos \omega t$
$$\begin{aligned} &\text { The path difference between the signal from } D \text { with that of } A \text { is }\\ &\begin{aligned} \sqrt{d^2+\left(\frac{\lambda}{2}\right)^2}-(d-\lambda / 2) & =d\left(1+\frac{\lambda}{4 d^2}\right)^{1 / 2}-d+\frac{\lambda}{2} \\ & =d\left(1+\frac{\lambda^2}{8 d^2}\right)^{1 / 2}-d+\frac{\lambda}{2} \approx \frac{\lambda}{2} \quad(\because d \gg \lambda) \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { Therefore, phase difference is } \pi \text {. }\\ &\begin{aligned} & \therefore \quad Y_D=a \cos (\omega t-\pi)=-a \cos \omega t \\ & \text { Thus, the signal picked up at } R_1 \text { from all the four sources is } Y_{R_1}=y_A+y_B+y_C+y_D \\ & \qquad=a \cos \omega t-a \cos \omega t+a \cos \omega t-a \cos \omega t=0 \end{aligned} \end{aligned}$$
(i) Let the signal picked up at $R_2$ from $B$ be $y_B=a_1 \cos \omega t$.
The path difference between signal at $D$ and that at $B$ is $\lambda / 2$.
$\therefore \quad y_D=-a_1 \cos \omega t$
The path difference between signal at $A$ and that at $B$ is
$$\sqrt{(d)^2+\left(\frac{\lambda}{2}\right)^2}-d=d\left(1+\frac{\lambda^2}{4 d^2}\right)^{1 / 2}-d \simeq \frac{1 \lambda^2}{8 d^2}$$
As $d \gg \lambda$, therefore this path difference $\rightarrow 0$
and phase difference $=\frac{2 \pi}{\lambda}\left(\frac{1}{8} \frac{\lambda^2}{d^2}\right) \rightarrow 0$
Hence, $y_A=a_1 \cos (\omega t-\phi)$
Similarly, $y_C=a_1 \cos (\omega t-\phi)$
$$\begin{aligned} &\therefore \text { Signal picked up by } R_2 \text { is }\\ &\begin{aligned} & y_A+y_B+y_C+y_D =y=2 a_1 \cos (\omega t-\phi) \\ \therefore & |y|^2 =4 a_1^2 \cos ^2(\omega t-\phi) \\ \therefore & < I > =2 a_1^2 \end{aligned} \end{aligned}$$
Thus, R$_1$ picks up the larger signal.
(ii) If $B$ is switched off, $R_1$ picks up
$$y=a \cos \omega t$$
$$\begin{aligned} \therefore & \left\langle I_{R_1}\right\rangle =\frac{1}{2} a^2 \\ R_2 \text { picks up } & y =a \cos \omega t \\ \therefore & \left\langle I_{R_2}\right\rangle =a^2<\cos ^2 \omega t>=\frac{a^2}{2} \end{aligned}$$
(iii) Thus, $R_1$ and $R_2$ pick up the same signal.
If $D$ is switched off.
$R_1$ picks up $y=a \cos \omega t$
$$\begin{array}{rlrl} \therefore & \left\langle I_{R_1}\right\rangle =\frac{1}{2} a^2 \\ & R_2 \text { picks up } & y =3 a \cos \omega t \\ \therefore & \left\langle I_{R_2}\right\rangle =9 a^2<\cos ^2 \omega t>=\frac{9 a^2}{2} \end{array}$$
(iii) Thus, $R_1$ and $R_2$ pick up the same signal.
If $D$ is switched off.
$R_1$ picks up $y=a \cos \omega t$
$\therefore \quad\left\langle I_{R_1}\right\rangle=\frac{1}{2} a^2$
$R_2$ picks up $$y=3 a \cos \omega t$$
$$\therefore \quad\left\langle I_{R_2}\right\rangle=9 a^2<\cos ^2 \omega t>=\frac{9 a^2}{2}$$
Thus, $R_2$ picks up larger signal compared to $R_1$.
(iv) Thus, a signal at $R_1$ indicates $B$ has been switched off and an enhanced signal at $R_2$ indicates $D$ has been switched off.
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, $AB=2d\tan r\sin i.$
$$\begin{aligned} \text { Thus, the optical path difference } & =\frac{2 n d}{\cos r}-2 d \tan r \sin i \\ & =2 \cdot \frac{\sin i d}{\sin r \cos r}-2 d \frac{\sin r}{\cos r} \sin i \\ & =2 d \sin \left[\frac{1-\sin ^2 r}{\sin r \cos r}\right] \\ & =2 n d \cos r \end{aligned}$$
$$\begin{aligned} & \text { For these waves to interfere destructively path difference }=\frac{\lambda}{2} \text {. } \\ & \Rightarrow \quad 2 n d \cos r=\frac{\lambda}{2} \\ & \Rightarrow \quad n d \cos r=\frac{\lambda}{4}\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} &\text { For photographic lenses, the sources are normally in vertical plane }\\ &\begin{aligned} & \therefore \quad i=r=0 \Upsilon \\ & \text { From Eq. (i), } \quad n d \cos 0 \Upsilon=\frac{\lambda}{4} \\ & \Rightarrow \quad d=\frac{\lambda}{4 n} \\ & =\frac{5500 \mathop A\limits^o}{4 \times 1.38} \approx 1000 \mathop A\limits^o \end{aligned} \end{aligned}$$