When angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.

Polarisation by reflection occurs when the angle of incidence is the Brewster's angle i.e.,

$$\tan i_B={ }^1 \alpha_2=\frac{\alpha_2}{\alpha_1} \text { where } \alpha_2<\alpha_1$$

when the light rays travels in such a medium, the critical angle is

$$\sin i_c=\frac{\alpha_2}{\alpha_1}$$

where, $\alpha_2<\alpha_1$

As $\left|\tan i_B\right|>\left|\sin i_C\right|$ for large angles $i_B

Thus, the polarisation by reflection occurs definitely.

We know that

$$\text { Resolving power }=\frac{1}{d}=\frac{2 \sin \beta}{1.22 \lambda} \Rightarrow d_{\min }=\frac{1.22 \lambda}{2 \sin \beta}$$

where, $\lambda$ is the wavelength of light and $\beta$ is the angle subtended by the objective at the object.

For the light of wavelength $5500 \mathop A\limits^o$,

$$d_{\min }=\frac{1.22 \times 5500 \times 10^{-10}}{2 \sin \beta}\quad\text{.... (i)}$$

For electrons accelerated through 100 V , the de-Broglie wavelength

$$\begin{aligned} \lambda & =\frac{12.27}{\sqrt{V}}=\frac{12.27}{\sqrt{100}}=0.12 \times 10^{-9} \mathrm{~m} \\ d_{\min } & =\frac{1.22 \times 0.12 \times 10^{-9}}{2 \sin \beta} \end{aligned}$$

Ratio of the least separation

$$\therefore \quad \frac{d_{\min }^{\prime}}{d_{\min }}=\frac{0.12 \times 10^{-9}}{5500 \times 10^{-10}}=0.2 \times 10^{-3}$$

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

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