Will the focal length of a lens for red light be more, same or less than that for blue light?
As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
In other words, $\propto_b>\propto_r$
By lens maker's formula,
$$\frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$
Therefore, $f_b< f_t$.
Thus, the focal length for blue light will be smaller than that for red.
The near vision of an average person is 25 cm . To view an object with an angular magnification of 10 , what should be the power of the microscope?
The least distance of distinct vision of an average person (i.e., $D$ ) is 25 cm , in order to view an object with magnification 10,
Here, $v=D=25 \mathrm{~cm}$ and $u=f$
But the magnification $m=v / u=D / f$
$$\begin{aligned} & m=\frac{D}{f} \\ \Rightarrow\quad & f=\frac{D}{m}=\frac{25}{10}=2.5=0.025 \mathrm{~m} \\ & P=\frac{1}{0.025}=40 \mathrm{D} \end{aligned}$$
This is the required power of lens.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?
No, the reversibility of the lens maker's equation.
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.