From the given figure of two slit interference arrangements, we can write

$$\begin{aligned} & T_2 P=T_2 O+O P=D+x \\ \text{and}\quad & T_1 P=T_1 O-O P=D-x \\ & S_1 P=\sqrt{\left(S_1 T_1\right)^2+\left(P T_1\right)^2}=\sqrt{D^2+(D-x)^2} \\ \text{and}\quad & S_2 P=\sqrt{\left(S_2 T_2\right)^2+\left(T_2 P\right)^2}=\sqrt{D^2+(D+x)^2} \end{aligned}$$

The minima will occur when $S_2 P-S_1 P=(2 n-1) \frac{\lambda}{2}$

i.e., $\quad\left[D^2+(D+x)^2\right]^{1 / 2}-\left[D^2+(D-x)^2\right]^{1 / 2}=\frac{\lambda}{2}\quad$ [for first minima $n=1$ ]

If $x=D$

$$ \begin{aligned} & \text { we can write } \\ & {\left[D^2+4 D^2\right]^{1 / 2}-\left[D^2+0\right]^{1 / 2}=\frac{\lambda}{2}} \\ & \Rightarrow \quad\left[5 D^2\right]^{1 / 2}-\left[D^2\right]^{1 / 2}=\frac{\lambda}{2} \\ & \Rightarrow \quad \sqrt{5} D-D=\frac{\lambda}{2} \\ & \Rightarrow \quad D(\sqrt{5}-1)=\lambda / 2 \text { or } D=\frac{\lambda}{2(\sqrt{5}-1)} \\ \text{Putting}\quad & \sqrt{5}=2.236 \\ \Rightarrow\quad & \sqrt{5}-1=2.236-1=1.236 \\ & D=\frac{\lambda}{2(1.236)}=0.404 \lambda \end{aligned}$$

$$\begin{aligned} & A=\text { Resultant amplitude } \\ & =A \text { parallel }\left(A_{\|}\right)+A \text { perpendicular }\left(A_{\perp}\right) \\ & \Rightarrow \quad A=A_{\perp}+A_{\|} \\ & \text {Without } P \\ & A=A_{\perp}+A_{\|} \end{aligned}$$

$$\begin{aligned} A_1=A_{\perp}^1+A_{\perp}^2 & =A_{\perp}^0 \sin (k x-\omega t)+A_{\perp}^0 \sin (k x-\omega t+\phi) \\ A_{\|} & =A_{\|}^{(1)}+A_{\|}^{(2)} \\ A_{\|} & =A_{\|}^0[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)] \end{aligned}$$

where $A_{\perp}^0, A_{\|}^0$ are the amplitudes of either of the beam in perpendicular and parallel polarisations.

$$\therefore \text { Intensity }=\left\{\left|A_{\perp}^0\right|^2+\left|A_{\|}^0\right|^2\right\}\left[\sin ^2(k x-\omega t)\left(1+\cos ^2 \phi+2 \sin \phi\right)+\sin ^2(k x-\omega t) \sin ^2 \phi\right]$$

$$\begin{aligned} & =\left\{\left|A_{\perp}^0\right|^2+\left|A_{\| \mid}^0\right|^2\right\}\left(\frac{1}{2}\right) \cdot 2(1+\cos \phi) \\ & =2\left|A_{\perp}^0\right|^2(1+\cos \phi), \text { since, }\left|A_{\perp}^0\right|_{\mathrm{av}}=\left|A_{\|}^0\right|_{\mathrm{av}} \end{aligned}$$

With $P$

Assume $A_{\perp}^2$ is blocked

$$\begin{aligned} \text { Intensity } & =\left(A_{\|}^1+A_{\|}^2\right)^2+\left(A_{\perp}^1\right)^2 \\ & =\left|A_{\perp}^0\right|^2(1+\cos \phi)+\left|A_{\perp}^0\right|^2 \cdot \frac{1}{2} \end{aligned}$$

Given, $I_0=4\left|A_{\perp}^0\right|^2=$ Intensity without polariser at principal maxima.

Intensity at principal maxima with polariser

$$=\left|A_{\perp}^0\right|^2\left(2+\frac{1}{2}\right)=\frac{5}{8} I_0$$

Intensity at first minima with polariser

$$=\left|A_{\perp}^0\right|^2(1-1)+\frac{\left|A_{\perp}^0\right|^2}{2}=\frac{I_0}{8} .$$

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

$$D \tan \theta^{\prime \prime}{ }_1=\frac{5 D}{\sqrt{16^2-5^2}}=\frac{5 D}{\sqrt{231}} \text { below point } O.$$

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.

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.