Three immiscible liquids of densities $d_1>d_2>d_3$ and refractive indices $\propto_1>\propto_2>\propto_3$ are put in a beaker. The height of each liquid column is $\frac{h}{3}$. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.
$$\begin{aligned} &\text { Let the apparent depth be } O_1 \text { for the object seen from } m_2 \text {, then }\\ &O_1=\frac{\propto_2}{\propto_1} \frac{h}{3} \end{aligned}$$
Since, apparent depth $=$ real depth $/$ refractive index $\propto$.
Since, the image formed by Medium $1, \mathrm{O}_2$ act as an object for Medium 2.
If seen from $\propto_3$, the apparent depth is $\mathrm{O}_2$.
Similarly, the image formed by Medium $2, \mathrm{O}_2$ act as an object for Medium 3
$$ \begin{aligned} O_2 & =\frac{\propto_3}{\propto_2}\left(\frac{h}{3}+O_1\right) \\ & =\frac{\propto_3}{\propto_2}\left(\frac{h}{3}+\frac{\propto_2 h}{\propto_1 3}\right)=\frac{h}{3}\left(\frac{\propto_3}{\propto_2}+\frac{\propto_2}{\propto_1}\right) \end{aligned}$$
Seen from outside, the apparent height is
$$\begin{aligned} O_3 & =\frac{1}{\propto_3}\left(\frac{h}{3}+O_2\right)=\frac{1}{\propto_3}\left[\frac{h}{3}+\frac{h}{3}\left(\frac{\propto_3}{\propto_2}+\frac{\alpha_3}{\propto_1}\right)\right] \\ & =\frac{h}{3}\left(\frac{1}{\propto_1}+\frac{1}{\propto_2}+\frac{1}{\propto_3}\right) \end{aligned}$$
This is the required expression of apparent depth.
For a glass prism $(\propto=\sqrt{3})$, the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.
The relationship between refractive index, prism angle $A$ and angle of minimum deviation is given by
$$\propto=\frac{\sin \left[\frac{\left(A+D_m\right)}{2}\right]}{\sin \left(\frac{A}{2}\right)}$$
Here,
$\therefore$ Given, $$D_m=A$$
Substituting the value, we have
$$\begin{aligned} & \therefore \quad \alpha=\frac{\sin A}{\sin \frac{A}{2}} \\ & \text { On solving, we have } \quad =\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2} \end{aligned}$$
$$\propto=\frac{\sin A}{\sin \frac{A}{2}}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}=2 \cos \frac{A}{2}$$
For the given value of refractive index, we have
$$\begin{array}{llr} \therefore & \cos \frac{A}{2}=\frac{\sqrt{3}}{2} \\ \text { or } & \frac{A}{2} =30 \Upsilon \\ \therefore & A =60 \Upsilon \end{array}$$
This is the required value of prism angle.
A short object of length $L$ is placed along the principal axis of a concave mirror away from focus. The object distance is $u$. If the mirror has a focal length $f$, what will be the length of the image? You may take $L<|v-f|$.
Since, the object distance is $u$. Let us consider the two ends of the object be at distance $u_1=u-L / 2$ and $u_2=u+L / 2$, respectively so that $\left|u_1-u_2\right|=L$. Let the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length would be
$L^{\prime}=\left|v_1-v_2\right|\quad\text{.... (i)}$
Applying mirror formula, we have
$$\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \text { or } v=\frac{f u}{u-f}$$
On solving, the positions of two images are given by
$$v_1=\frac{f(u-L / 2)}{u-f-L / 2}, v_2=\frac{f(u+L / 2)}{u-f+L / 2}$$
For length, substituting the value in (i), we have
$$L^{\prime}=\left|v_1-v_2\right|=\frac{f^2 L}{(u-f)^2 \times L^2 / 4}$$
Since, the object is short and kept away from focus, we have
$$\begin{gathered} L^2 / 4<<(u-f)^2 \\ \text{Hence, finally}\quad L^{\prime}=\frac{f^2}{(u-f)^2} L \end{gathered}$$
This is the required expression of length of image.
A circular disc of radius $R$ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius a figure. The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index $\propto$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?
Refering to the figure, $A M$ is the direction of incidence ray before liquid is filled. After liquid is filled in , $B M$ is the direction of the incident ray. Refracted ray in both cases is same as that along $A M$.
Let the disc is separated by $O$ at a distance $d$ as shown in figure. Also, considering angle
$$N=90 \Upsilon, O M=a, C B=N B=a-R, A N=a+R$$
Here, in figure
$$\begin{aligned} \sin t & =\frac{a-R}{\sqrt{d^2+(a-R)^2}} \\ \text{and}\quad\sin \alpha & =\cos (90-\alpha)=\frac{a+R}{\sqrt{d^2+(a+R)^2}} \end{aligned}$$
But on applying Snell's law,
$$\frac{1}{\alpha}=\frac{\sin t}{\sin r}=\frac{\sin t}{\sin \alpha}$$
On substituting the values, we have the separation
$$d=\frac{\alpha\left(a^2-b^2\right)}{\sqrt{(a+r)^2-\alpha(a-r)^2}}$$
This is the required expression.
A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at $(0,0)$ and an object placed at $(-50 \mathrm{~cm}, 0)$. Find the coordinates of the image.
If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO'.
Applying lens formula, we have
$$\begin{aligned} \frac{1}{v}-\frac{1}{u} & =\frac{1}{f} \\ \frac{1}{v} & =\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} \\ v & =50 \mathrm{~cm} \\ \text { Mangnification is } m & =\frac{v}{u}=-\frac{50}{50}=-1 \end{aligned}$$
Thus, the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence, with respect to the $X$-axis passing through the edge of the cut lens, the coordinates of the image are $(50 \mathrm{~cm},-1 \mathrm{~cm})$.