'For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000 \mathop A\limits^o$ and electrons accelerated through 100 V used as the illuminating substance.
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
$$\therefore \quad \frac{d_{\min }^{\prime}}{d_{\min }}=\frac{0.12 \times 10^{-9}}{5500 \times 10^{-10}}=0.2 \times 10^{-3}$$
Consider a two slit interference arrangements (figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre 0.
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 $x=D$
$$ \begin{aligned} & \text { we can write } \\ & {\left[D^2+4 D^2\right]^{1 / 2}-\left[D^2+0\right]^{1 / 2}=\frac{\lambda}{2}} \\ & \Rightarrow \quad\left[5 D^2\right]^{1 / 2}-\left[D^2\right]^{1 / 2}=\frac{\lambda}{2} \\ & \Rightarrow \quad \sqrt{5} D-D=\frac{\lambda}{2} \\ & \Rightarrow \quad D(\sqrt{5}-1)=\lambda / 2 \text { or } D=\frac{\lambda}{2(\sqrt{5}-1)} \\ \text{Putting}\quad & \sqrt{5}=2.236 \\ \Rightarrow\quad & \sqrt{5}-1=2.236-1=1.236 \\ & D=\frac{\lambda}{2(1.236)}=0.404 \lambda \end{aligned}$$
Figure shown a two slit arrangement with a source which emits unpolarised light. $P$ is a polariser with axis whose direction is not given. If $I_0$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
$$\begin{aligned} & A=\text { Resultant amplitude } \\ & =A \text { parallel }\left(A_{\|}\right)+A \text { perpendicular }\left(A_{\perp}\right) \\ & \Rightarrow \quad A=A_{\perp}+A_{\|} \\ & \text {Without } P \\ & A=A_{\perp}+A_{\|} \end{aligned}$$
$$\begin{aligned} A_1=A_{\perp}^1+A_{\perp}^2 & =A_{\perp}^0 \sin (k x-\omega t)+A_{\perp}^0 \sin (k x-\omega t+\phi) \\ A_{\|} & =A_{\|}^{(1)}+A_{\|}^{(2)} \\ A_{\|} & =A_{\|}^0[\sin (k x-\omega t)+\sin (k x-\omega t+\phi)] \end{aligned}$$
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
$$=\left|A_{\perp}^0\right|^2(1-1)+\frac{\left|A_{\perp}^0\right|^2}{2}=\frac{I_0}{8} .$$
$$A C=C O=D, S_1 C=S_2 C=d \ll D$$
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.