Ray Optics

For the end $B$, image will be formed at,

$ \begin{aligned} & \frac{1}{v_1}=\frac{1}{f}-\frac{1}{u} \\ & \frac{1}{v_1}=\frac{1}{-9}=\frac{1}{-15}=\frac{-2}{45} \mathrm{~cm} \\ & v_1=-22.5 \mathrm{~cm} \end{aligned} $

For end $A$ of rod, image will be formed at

$ \begin{aligned} & \frac{1}{v_2}=\frac{1}{f}-\frac{1}{u} \\ & \frac{1}{v_2}=\frac{1}{-9}-\frac{1}{-21}=-\frac{4}{63} \mathrm{~cm} \\ & v_2=-15.75 \mathrm{~cm} \end{aligned} $

∴ Length of image $=\left|v_1-v_2\right|=6.75 \mathrm{~cm}$

$ \begin{aligned} &\text { Applying Snell's law at first surface }\\ &\begin{aligned} & 1 \sin 9.3^{\circ}=\mu \sin 6^{\circ} \quad\left[\because r_1=A\right] \\ & \mu=\frac{\sin 9.3}{\sin 6^{\circ}}=\frac{9.3}{6}=1.55 \end{aligned} \end{aligned} $

From the figure,

$ \begin{aligned} &\begin{aligned} & u=-(f+16) \\ & v=-(f+9) \end{aligned}\\ &\text { Using mirror formula }\\ &\begin{array}{ll} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ & -\frac{1}{f+9}-\frac{1}{f+16}=-\frac{1}{f} \\ \Rightarrow & \frac{-f-16-f-9}{(f+9)(f+16)}=\frac{-1}{f} \\ \Rightarrow & +2 f^2+25 f=f^2+25 f+144 \\ \Rightarrow & f^2=144 \\ \Rightarrow & f=12 \mathrm{~cm} \end{array} \end{aligned} $

$ \begin{aligned} &\text { For equilateral prism, } A=60^{\circ}\\ &\begin{aligned} \therefore \quad \delta_m & =A=60^{\circ} \\ \mu & =\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \frac{A}{2}} \\ & =\frac{\sin \left(\frac{60+60}{2}\right)}{\sin \frac{60}{2}}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}} \end{aligned} \end{aligned} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}=1.732 $

Focal length of primary mirror

$ f_1=\frac{R_1}{2}=\frac{25}{2}=12.5 \mathrm{~cm} $

Since, object is at infinity, so the image formed by the primary mirror will be at its focus.

$ \therefore \quad v_1=f_1=12.5 \mathrm{~cm} $

Focal length of secondary mirror,

$ \begin{array}{ll} & f_2=-\frac{R_2}{2}=\frac{-16}{2}=-8 \mathrm{~cm} \\ \therefore \quad & u=12.5-2.5=10 \mathrm{~cm} \end{array} $

Using mirror formula,

$ \begin{aligned} \frac{1}{f} & =\frac{1}{v}+\frac{1}{u} \\ \frac{1}{-8} & =\frac{1}{v}+\frac{1}{-10} \Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{8}=-\frac{1}{40} \\ \Rightarrow \quad v & =-40 \mathrm{~cm} \end{aligned} $

Thus, the final image is formed at a distance of 40 cm from the convex mirror.

$ \text { } \begin{aligned} \frac{1}{f} & =\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & =\left(\frac{1.5}{1.3}-1\right)\left(\frac{1}{6}-\frac{1}{-12}\right) \\ \frac{1}{f} & =\frac{0.2}{1.3} \times \frac{3}{12} \\ f & =\frac{1.3 \times 12}{0.2 \times 3}=13 \times 2=26 \mathrm{~cm} \end{aligned} $

Step 1: Using Brewster's Law

Brewster's law says that $\mu = \tan i_p$, where $\mu$ is the refractive index of the medium, and $i_p$ is the polarising angle (angle of incidence for which the reflected light is completely polarised).

We know the refractive index $\mu = \sqrt{3}$. So, $\sqrt{3} = \tan i_p$.

We know that $\tan 60^\circ = \sqrt{3}$. So, $i_p = 60^\circ$.

Step 2: Finding the Angle of Refraction

When light is totally polarised after reflection, the angle between the reflected and refracted rays is $90^\circ$. So, $i_p + r = 90^\circ$, where $r$ is the angle of refraction.

This means $r = 90^\circ - i_p$.

We already found $i_p = 60^\circ$, so $r = 90^\circ - 60^\circ = 30^\circ$.

Focal length of concave mirror

$ f=\frac{R}{2}=\frac{-30}{2}=-15 \mathrm{~cm} $

Object distance from the pole of the mirror

$ u=-(f+10)=-(-15+10)=5 \mathrm{~cm} $

According to question, for real image formed by a concave mirror, the object must be placed beyond the focal point. If the object is 10 cm from the principle focus and the focal length is 15 cm , the object distance from the pole is

$ u=-(15+10)=-25 \mathrm{~cm} $

Using mirror formula,

$ \begin{aligned} & \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \Rightarrow & \frac{1}{-15}=\frac{1}{-25}+\frac{1}{v} \\ \Rightarrow & \frac{1}{v}=\frac{1}{25}-\frac{1}{15}=\frac{-2}{75} \\ \therefore & v=-37.5 \mathrm{~cm} \\ \therefore & m=\frac{h_i}{h_0}=\frac{-v}{u} \\ \Rightarrow & \frac{h_i}{3.6}=\frac{-37.5}{-25} \\ \Rightarrow & h_i=-5.4 \mathrm{~cm} \end{aligned} $

The negative sign indicates that the image is real and inverted.

To find the angle of dispersion, we use the formula:

$\delta = (n_1 - n_2)A$

Here, $ n_1 $ and $ n_2 $ are the refractive indices for violet and red lights, and $ A $ is the prism's angle.

Plug in the values: $ n_1 = 1.52 $, $ n_2 = 1.48 $, $ A = 6^\circ $.

$\delta = (1.52 - 1.48) \times 6 = 0.04 \times 6 = 0.24^\circ$

So, the angle of dispersion is $ 0.24^\circ $, which is about $ 0.2^\circ $.

Given, the focal length of the objective lens is $f_0=1.25 \mathrm{~cm}$

The focal length of the eyepiece is

$ f_e=5 \mathrm{~cm} $

The separation between the objective and eyepiece is $L=7.5 \mathrm{~cm}$

The distance of this image from the objective lens is

$ \begin{aligned} & v_0=L-f_e \\ & v_0=7.5 \mathrm{~cm}-5 \mathrm{~cm}=2.5 \mathrm{~cm} \end{aligned} $

The lens formula for the objective lens

$ \begin{aligned} & \frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0} \\ & \frac{1}{u_0}=\frac{1}{2.5 \mathrm{~cm}}-\frac{1}{1.25 \mathrm{~cm}} \\ & u_0=-2.5 \mathrm{~cm} \end{aligned} $

The magnification of the objective lens is

$ M_0=-\frac{v_0}{u_0}=\frac{-2.5 \mathrm{~cm}}{-2.5 \mathrm{~cm}}=1 $

The magnification of the eyepiece when the final image is at infinity is $M_e=\frac{D}{f_e}$

$ M_e=\frac{25 \mathrm{~cm}}{5 \mathrm{~cm}}=5 $

The total magnification is $M=M_0 \times M_e$

$ M=1 \times 5=5 $

Dispersion is the property of light that causes the separation of light into its component colours when passing through a medium, such as thick lens. This happens because different colours(wavelengths) of light are refracted by different amount.

$\mu=\sqrt{2}, A=90^{\circ}$

For total internal reflection

$ \begin{aligned} & & \sin i_C & =\frac{1}{\mu}=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \\ & \therefore & i_C & =45^{\circ} \end{aligned} $

For total internal reflection to occur at the second face, the angle of incidence at the second face $\left(r_2\right)$ must be greater than or equal to the critical angle. To ensure total internal reflection, we consider the limiting case where

$ r_2=\theta_c=45^{\circ} $

The angle of the prism is related to the angles of refraction at the two faces by

$ A=r_1+r_2 $

Since $A=90^{\circ}$ and $r_2=45^{\circ}$

$ \begin{aligned} & 90^{\circ}=r_1+45^{\circ} \\ & r_1=90^{\circ}-45^{\circ}=45^{\circ} \end{aligned} $

Applying Snell's law at the first face,

$ \begin{gathered} \mu_{\text {air }} \sin i=\mu_{\text {prism }} \sin r_1 \\ 1 \cdot \sin i=\sqrt{2} \sin 45^{\circ} \\ \sin i=\sqrt{2} \cdot \frac{1}{\sqrt{2}} \\ =1=\sin 90^{\circ} \\ \therefore \quad i=90^{\circ} \end{gathered} $

When the final image is at the least distance of distinct vision,

$ M=M_0 \cdot M_e $

as we know, $M_e=1+\frac{D}{F_e}$

$ M_e=1+\frac{25}{5}=6 $

$ \begin{aligned}So,\,\, & 24=M_0 \times 6 \\ & M_0=4 \end{aligned} $

Since, $A=r_1+r_2$

$ \Rightarrow \quad 30=r_1+0 \Rightarrow r_1=30^{\circ} $

∴ At surface $A B$

$ \begin{aligned} & n_1 \sin i=n_2 \sin r_1 \\ \Rightarrow & 1 \cdot \sin 60^{\circ}=n_2 \sin 30^{\circ} \\ \Rightarrow \quad & \frac{\sqrt{3}}{2}=n_2 \frac{1}{2} \Rightarrow n_2=\sqrt{3} \end{aligned} $

$ \text { } \begin{aligned} h_i & =n h_0 \\ n & =\frac{h_l}{h_0} \\ n & =-\frac{v}{u} \\ \Rightarrow \quad v & =-n u \end{aligned} $

Using mirror formula,

$ \begin{aligned} \frac{1}{f} & =\frac{1}{v}+\frac{1}{u} \\ \frac{1}{f} & =\frac{1}{-n u}+\frac{1}{u} \\ \Rightarrow \quad u & =\frac{f(n-1)}{n}=f-\frac{f}{n} \end{aligned} $

By Lens maker's formula, For a equviconvex lens,

$ \frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{2}{R}\right) $

In air, $n_1=1, n_2=\frac{3}{2}$

$ \begin{aligned} \therefore \quad \frac{1}{f_{\text {air }}} & =\left(\frac{3}{2}-1\right)\left(\frac{2}{R}\right) \\ \frac{1}{f_{\text {air }}} & =\frac{1}{R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

In liquid, $n_1=x, n_2=\frac{3}{2}$

$ \frac{1}{f_l}=\left(\frac{3}{2 x}-1\right)\left(\frac{2}{R}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Given, $\frac{f_{\text {air }}}{f_{\text {liquid }}}=\frac{1}{2}$

So, from Eq. (i) and (ii)

$ \begin{aligned} & \frac{f_{\text {air }}}{f_{\text {liquid }}}=\frac{R \times 2\left(\frac{3}{2 x}-1\right)}{R \times 1}=\frac{1}{2} \\ \Rightarrow & 2\left(\frac{3}{2 x}-1\right)=\frac{1}{2} \\ \Rightarrow & \frac{3}{2 x}-1=\frac{1}{4} \\ \Rightarrow & \frac{3}{2 x}=\frac{5}{4} \text { or } x=\frac{6}{5}=1: 20 \end{aligned} $

$ \text { } \begin{aligned} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\ & =\frac{1}{-35}-\frac{1}{-25}=\frac{10}{875} \\ \therefore \quad f & =\frac{875}{10}=87.5 \mathrm{~cm} \end{aligned} $

For surface $X Y$,

$ \begin{aligned} \mu_a \sin 45^{\circ} & =\mu g \sin r_1 & \\ \mu_g \sin r_1 & =\frac{1}{\sqrt{2}} & \\ \Rightarrow \quad & \sin r_1 =\frac{1}{\mu_g \sqrt{2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$\therefore$ From figure, $i_2=90-r_1$

For total internal reflection

$ \begin{aligned} & \sin i_2=\frac{1}{\mu_g} \\ \Rightarrow & \sin \left(90-r_1\right)=\frac{1}{\mu_g} \\ \Rightarrow & \cos r_1=\frac{1}{\mu_g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From eqn. (i) and (ii),

$ \begin{aligned} & \sin ^2 r_1+\cos ^2 r_1=1 \\ \Rightarrow \quad & \frac{1}{2 \mu_g^2}+\frac{1}{\mu_g^2}=1 \Rightarrow \frac{3}{2 \mu_g^2}=1 \\ \Rightarrow \quad & \mu_g^2=\frac{3}{2} ; \therefore \mu_g=\sqrt{\frac{3}{2}} \end{aligned} $

Magnification of a lens is given as

$ m=\frac{v}{u} $

from lens formula

$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{v}=\frac{f+u}{f u} \\ \Rightarrow \quad & \frac{u}{v}=\frac{f+u}{f} \end{aligned} $

and  $ m=\frac{v}{u}=\frac{f}{f+u} $

Now, magnification in both case is equal.

So, $m_1=-m_2$

$ \begin{aligned} \frac{f}{f-20} & =\frac{-f}{f-10} \\ f-20 & =-f+10 \\ 2 f & =30 \Rightarrow f=15 \mathrm{~cm} \end{aligned} $

$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{60+\delta m}{2}\right)}{\sin \frac{60}{2}} \\ & \Rightarrow \sqrt{3} \sin 30^{\circ}=\sin \left(\frac{60+\delta m}{2}\right) \\ & \Rightarrow \frac{\sqrt{3}}{2}=\sin \left(\frac{60+\delta m}{2}\right) \\ & \Rightarrow 60^{\circ}=\frac{60+\delta m}{2} \Rightarrow \delta m=60^{\circ} \end{aligned} $

$\therefore$ Angle of incidence

$ i=\frac{A+\delta m}{2}=\frac{60+60}{2}=60^{\circ} $

$ \begin{aligned} & \text { } f_1+f_2=18 \mathrm{~cm} \\ & \qquad \begin{array}{c} f=4 \mathrm{~cm} \\ \therefore \quad \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2} \\ \Rightarrow \quad \frac{1}{4}=\frac{f_2+f_1}{f_1 f_2} \\ \Rightarrow \quad \frac{1}{4}=\frac{18}{f_1 f_2} \\ \Rightarrow \quad f_1 f_2=72 \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad f_1\left(18-f_1\right)=72 \\ & \Rightarrow \quad 18 f_1-f_1^2=72 \\ & \Rightarrow \quad f_1^2-18 f_1+72=0 \\ & \Rightarrow \quad f_1^2-12 f_1-6 f^{\prime}+72=0 \\ & \Rightarrow \quad\left(f_1-12\right)\left(f_1-6\right)=0 \\ & \therefore \quad f_1=6 \mathrm{~cm}, 12 \mathrm{~cm} \end{aligned} $

When, $f_1=6 \mathrm{~cm}$, then,

$ f_2=18-6=12 \mathrm{~cm} $

When $f_1=12 \mathrm{~cm}$, then

$ f_2=18-12=6 \mathrm{~cm} $

Since, $P \propto \frac{1}{f}$

$\therefore$ Focal length of low power $=12 \mathrm{~cm}$

The far point of a short-sighted person is 400 cm . To correct this, a lens is needed to bring objects at infinity into focus at the person's far point.

$ \therefore P=\frac{1}{f}=\frac{1}{-4}=-0.250 $

Given, for concave mirror

$ \begin{array}{l} u=-24 \mathrm{~cm} \\\\ v=-12 \mathrm{~cm} \end{array} $

Using the mirror equation,

$ \begin{array}{l} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\\\ \frac{1}{f}=\frac{1}{-12}-\frac{1}{24} \\\\ \frac{1}{f}=\frac{-2-1}{24} \\\\ f=-8 \mathrm{~cm} \end{array} $

Magnification $ m $ is given by,

$ \begin{array}{l} m=\frac{v}{u} \\\\ m=\frac{-12}{-24}=\frac{1}{2} \end{array} $

The speed of the image $ v_{i} $ is related to the speed of object $ v_{0} $ by the magnification.

$ v_{i}=m^{2} \cdot v_{0} $

$ \begin{array}{l} v_{i}=\left(\frac{1}{2}\right)^{2} \times 12 \\\\ v_{i}=\frac{12}{4} \Rightarrow v_{i}=3 \mathrm{~m} / \mathrm{s} \end{array} $

The image of the object can be located on the screen for two-positions of convex lens such that $ u $ and $ v $ are exchanged.

The separation between two positions of the lens is,

$d=30 \mathrm{~cm}$

From the figure,

$ \begin{array}{l} u_{1}+v_{1}=90 \mathrm{~cm} \\\\ v_{1}-u_{1}=30 \mathrm{~cm} \end{array} $

On solving Eq. (i) and Eq. (ii), we get

$ \begin{array}{l} v_{1}=60 \mathrm{~cm} \\\\ u_{1}=30 \mathrm{~cm} \end{array} $

From lens formula,

$ \begin{array}{l} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\\\ \frac{1}{f}=\frac{1}{60}-\frac{1}{-30} \\\\ \frac{1}{f}=\frac{1+2}{60} \Rightarrow f=20 \mathrm{~cm} \end{array} $

When monochromatic light strikes a surface that separates two different media, both the reflected and refracted light will have the same frequency as the original light.

This phenomenon occurs because the reflection and refraction processes involve the interaction of light with the atoms at the boundary. These atoms act like oscillators and are compelled to vibrate at the frequency of the incoming light. Consequently, the light emitted by these oscillating atoms—whether reflected or refracted—retains the same frequency as the incident light.

The power of a thin convex lens when placed in air is given as $ +4 \, \text{D} $, with the refractive index of the lens material being $\frac{3}{2}$. Let's explore how the lens behaves when immersed in a liquid with a refractive index of $\frac{5}{3}$.

Lens in Air

Power Formula:

$ P_{\text{air}} = \frac{1}{f} $

Here, the power $ P_{\text{air}} = 4 \, \text{D} $, thus:

$ f = \frac{1}{4} \, \text{m} = 25 \, \text{cm} $

Lens Maker's Formula in Air:

$ \frac{1}{f} = \left(\mu - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Substituting values:

$ \frac{1}{f} = \left(\frac{3}{2} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

$ 4 = \frac{1}{2}\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

$ \frac{1}{R_1} - \frac{1}{R_2} = 8 $

Lens in Liquid

When the lens is immersed in a liquid of refractive index $\frac{5}{3}$, the effective focal length $ f' $ changes:

Refractive Index Ratio:

$ \frac{\mu_g}{\mu_w} = \frac{\frac{3}{2}}{\frac{5}{3}} = \frac{9}{10} $

Modified Lens Maker's Formula in Liquid:

$ \frac{1}{f'} = \left(\frac{\mu_g}{\mu_w} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Substituting values:

$ \frac{1}{f'} = \left(\frac{9}{10} - 1\right) \times 8 $

$ \frac{1}{f'} = (-0.1) \times 8 = -0.8 $

New Focal Length:

$ f' = -\frac{1}{0.8} = -125 \, \text{cm} $

The negative focal length indicates that the lens behaves like a concave lens of focal length $125 \, \text{cm}$ when immersed in this liquid.

Given:

Refractive index, $ \mu = 1.6 $

Angle of minimum deviation, $ \delta_{D} = 4.2^{\circ} $

We wish to find the angle of the prism, denoted as $ A $.

The formula for the angle of minimum deviation is:

$ \delta_{D} = (\mu - 1) A $

Rearranging this formula to solve for $ A $, we have:

$ A = \frac{\delta_{D}}{\mu - 1} $

Substituting the known values:

$ A = \frac{4.2}{1.6 - 1} $

$ A = \frac{4.2}{0.6} $

Thus, the angle of the prism $ A $ is:

$ A = 7^{\circ} $

To determine the Brewster angle for the transition of light from air to glass, we start by noting the refractive index of glass, which is given as 1.5.

The Brewster angle $\theta$ is related to the refractive index $\mu$ through the following relationship:

$ \mu = \tan \theta $

Substituting the given refractive index into the equation:

$ 1.5 = \tan \theta $

This can also be expressed as:

$ \frac{3}{2} = \tan \theta $

Therefore, we find the Brewster angle by taking the inverse tangent:

$ \theta = \tan^{-1}\left(\frac{3}{2}\right) $

Given,

Object height, $ O=12 \mathrm{~cm} $

First image height, $ I_{1}=18 \mathrm{~cm} $

We know that product of magnification

$ m_{1} m_{2}=1 $

(where $ m_{1}= $ when image height is $ I_{1} $ and $ m_{2}= $ when image height is $ I_{2} $ )

$ \begin{aligned} m_{1} & =\frac{I_{1}}{0}=\frac{18}{12}=\frac{3}{2} \\\\ m_{1} m_{2} & =1 \\\\ \frac{3}{2} m_{2} & =1 \end{aligned} $

$ \Rightarrow \quad m_{2}=\frac{2}{3} $

Now, $ \quad m_{2}=\frac{I_{2}}{O} $

$ \begin{array}{l} \frac{2}{3}=\frac{I_{2}}{12} \\\\ I_{2}=\frac{2 \times 12}{3} \\\\ I_{2}=8 \mathrm{~cm} \end{array} $

New height of image.

Focal length of plano-convex

$ =73.5 \mathrm{~cm} $

Diameter of circular aperture $ =8.4 \mathrm{~cm} $

$ r=4.2 \mathrm{~cm} $

Refractive index, $ \mu=\frac{5}{3}=1.66 $

Radius of curvature and focal length relation is

$ \begin{aligned} f & =\frac{R}{\mu-1} \\\\ \Rightarrow \quad R & =f(\mu-1) \\\\ R & =73.5 \times 0.66 \\\\ R & =48.5 \end{aligned} $

Now using the relation

$ \begin{aligned} R & =\frac{r^{2}}{2 x}[x=\text { thickness of lens }] \\\\ 48.5 & =\frac{(4.2)^{2}}{2 x} \\\\ x & =\frac{17.64}{97} \mathrm{~cm} \\\\ x & =0.181 \mathrm{~cm} \\\\ x & =1.8 \mathrm{~mm} \end{aligned} $

Given, $\mu_{\text {glas }}=(3 / 2)$

As we know that, $\delta_m=(\mu-1) A$

Here, $A$ is prism angle

Now, $\delta_{m_2}=\left(\frac{3}{2}-1\right) A=\frac{1}{2} A$

$ \begin{aligned} & \delta_{m_2}=\left(\frac{\frac{3}{2}}{\frac{4}{3}}-1\right) A=\frac{1}{8} A \\\\ \Rightarrow & \frac{\delta_{m_m}}{\delta_{m_2}}=\frac{\frac{1}{2}}{\frac{1}{8}}=\frac{1}{2} \times 8=4: 1 \end{aligned} $

Given, power of lens $=2 \mathrm{D}$

Focal length $=\frac{1}{2} \mathrm{~m}$

$ =0.5 \mathrm{~m}=50 \mathrm{~cm} $

Hence, near point will also be 50 cm .

Given that,

Lens 1 has focal length $=f_1$

Lens 2 has focal length $=f_2$,

Here, when two convex lenses are combine to act as a glass \$lab, the distance of separation between them is the sum of their focal length, so the correct representation of the formula is $f_1+f_2$.

For an equilateral prism, the angle of the prism $(A)$ is $60^{\circ}$.

If angles of incidence ( $i$ ) and emergence

(e) are both $70 \%$ of $A$, then

$ i=e=0.7 \times A=0.7 \times 60^{\circ}=42^{\circ} $

The angle of minimum deviation $\left(\delta_m\right)$ occurs when the light path inside the prism is parallel to the base, which implies that $i=e$

Since, $i=e$, we have

$ \delta_m=i+e-A $

$ \begin{aligned} & =42^{\circ}+42^{\circ}-60^{\circ} \\\\ & =84^{\circ}-60^{\circ} \\\\ \delta_m & =24^{\circ} \end{aligned} $

The angle of minimum deviation is $24^{\circ}$.

Given, two different liquids of refractive indices 1.25 and 1.5 .

Focal length, $\frac{f_1}{f_2}=\frac{5}{16}$

Using len's Maker's formula,

$ \begin{aligned} & \frac{1}{f_1}=\frac{\left(n_{\text {Lens }}-1.25\right)}{1.25}\left(\frac{1}{R_1}-\frac{1}{R_{2^{\prime}}}\right) \ldots \\ & \frac{1}{f_2}=\frac{\left(n_{\text {Lens }}-1.5\right)}{1.5}\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \ldots \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned} & \frac{\left(\frac{1}{f_1}\right)}{\left(\frac{1}{f_2}\right)}=\frac{\left(n_{\text {Lens }}-1.25\right) \times 1.5}{\left(n_{\text {Lens }}-1.5\right) \times 1.25} \\ \Rightarrow \quad & \frac{16}{5}=\frac{\left(n_{\text {Lens }}-1.25\right)}{\left(n_{\text {Lens }}-1.5\right)} \times \frac{6}{5} \\ \Rightarrow \quad & 16 n_{\text {Lens }}-24=6 n_{\text {Lens }}-7.5 \\ \Rightarrow \quad & 10 n_{\text {Lens }}=16.5 \\ \Rightarrow \quad & n_{\text {Lens }}=1.65 \end{aligned} $

Therefore, the refractive index of material of the lens is approximately 1.65.

Given:

Let the refractive index of the lens be $\mu_L$.

The refractive index of the liquid is given by $0.8 \mu_L$.

Let's assign:

The focal length of the lens in air as $f$.

The focal length in the liquid as $f_{\text{liquid}} = 2f$.

Using the Lens Maker's formula in air:

$ \frac{1}{f} = (\mu_L - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \tag{i} $

For the lens immersed in the liquid, the formula modifies to:

$ \frac{1}{f_{\text{liquid}}} = \left(\frac{\mu_L}{\mu_{\text{liquid}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Substituting the given values:

$ \frac{1}{2f} = \left(\frac{\mu_L}{0.8\mu_L} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

This simplifies to:

$ \frac{1}{2f} = 0.25\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Therefore:

$ \Rightarrow \quad \frac{1}{f} = 0.5\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \tag{ii} $

From equations (i) and (ii), we derive:

$ \mu_L - 1 = 0.5 \quad \Rightarrow \quad \mu_L = 1.5 $

Thus, the refractive index of the liquid:

$ = 0.8 \times \mu_L = 0.8 \times 1.5 = 1.2 $

Therefore, the refractive index of the liquid is 1.2.

A person can see objects clearly when they are between 40 cm and 400 cm from their eyes. The goal is to increase the maximum distance of distinct vision to infinity.

The maximum distance at which the person can see clearly is 400 cm. To achieve clear vision at infinity, an object's image at infinity needs to be formed at 400 cm, or 4 meters.

Using the lens formula:

$ P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $

For this situation:

$ P = \frac{1}{-4} - \frac{1}{\infty} = \frac{1}{f} $

$ P = \frac{1}{f} = -0.25 \, \text{D} $

The negative sign indicates that a concave lens is required.

Given:

Focal length of the objective lens, $ f_0 = 30 \, \text{cm} $

Focal length of the eye lens, $ f_e = 3 \, \text{cm} $

Distance to the object being focused on, $ u_0 = -200 \, \text{cm} $ (since the object is 2 m away and the convention is to take the object distance as negative)

Let's use the lens formula to find the image distance ($ v_0 $) formed by the objective lens:

$ \frac{1}{f_0} = \frac{1}{v_0} - \frac{1}{u_0} $

Substitute the known values into the formula:

$ \frac{1}{30} = \frac{1}{v_0} + \frac{1}{200} $

Solving for $ \frac{1}{v_0} $:

$ \frac{1}{v_0} = \frac{1}{30} - \frac{1}{200} = \frac{20 - 3}{600} = \frac{17}{600} $

Thus, $ v_0 $ is:

$ v_0 = \frac{600}{17} \approx 35.3 \, \text{cm} $

The distance from the objective lens to the eye lens ($ d $) is the sum of $ v_0 $ and the focal length of the eye lens ($ f_e $):

$ d = v_0 + f_e = 35.3 + 3 = 38.3 \, \text{cm} $

Therefore, to see a clear image, the distance between the objective lens and the eye lens should be approximately 38.3 cm.

When a ray of light passes from an optically denser to rarer medium, the deviation is, $\delta=r-i$, where $i=$ angle of incidence $r=$ angle of refraction This can have a maximum value of $\left(\frac{\pi}{2}-C\right)$ for $i=C$ and $r=\frac{\pi}{2}$.

When total internal reflection occurs, the deviation is

$ \delta=\pi-2 i $

the minimum value of $i$ being $C$, then maximum value of

$ \delta_{\max }=\pi-2 C $

Given:

The angle of polarization for a medium is $\theta_p = 60^{\circ}$.

We know that:

$ \mu = \tan(\theta_p) \Rightarrow \mu = \tan 60^{\circ} = \sqrt{3} $

Furthermore, since the refractive index $\mu$ is also related to the critical angle by:

$ \mu = \frac{1}{\sin C} \quad \text{(where $C$ is the critical angle)} $

We can substitute for $\mu$:

$ \sqrt{3} = \frac{1}{\sin C} \Rightarrow \sin C = \frac{1}{\sqrt{3}} $

Therefore, the critical angle $C$ is:

$ C = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) $

Given the object distance $ u = -18 \, \text{cm} $ and the image distance $ v = 4 \, \text{cm} $, we need to find the focal length $ f $, the type of mirror, and the nature of the image.

Using the mirror formula:

$ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} $

Substitute the given values:

$ \frac{1}{f} = \frac{1}{-18} + \frac{1}{4} $

Solving for $ f $:

$ f = \frac{36}{7} $

$ f = 5.14 \, \text{cm} $

Since the focal length is positive ($ f > 0 $), the mirror is a convex mirror. Therefore, the image formed is virtual.

To determine the apparent depth of an object submerged in water and viewed from the air, we can use the concept of refractive index.

The refractive index ($\mu$) is defined by the equation:

$ \mu = \frac{\text{Real depth}}{\text{Apparent depth}} $

Given:

Refractive index of water, $\mu = \frac{4}{3}$

Real depth = 100 cm = 100 \times 10^{-2} m

We need to find the apparent depth. Plug the values into the formula:

$ \frac{4}{3} = \frac{100 \times 10^{-2}}{\text{Apparent depth}} $

Solve for the apparent depth:

$ \text{Apparent depth} = \frac{3 \times 100 \times 10^{-2}}{4} = 75.0 \times 10^{-2} m $

Thus, the apparent depth is 75 cm.

To find the refractive index of the liquid in the hollow prism, we consider the scenario of minimum deviation. Here's how we can calculate it:

Angular Relations:

At minimum deviation, the angle of the prism $ A $ is related to the angle of refraction $ r $ by:

$ A = 2r = 2 \times 30^\circ = 60^\circ $

Given: Minimum deviation $ \delta_m = 30^\circ $.

Refractive Index Calculation:

The refractive index $ \mu $ of the liquid is given by the formula:

$ \mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \frac{A}{2}} $

Substitute the known values:

$ \mu = \frac{\sin \left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin \frac{60^\circ}{2}} $

Simplify the expression:

$ \mu = \frac{\sin 45^\circ}{\sin 30^\circ} $

Trigonometric Values:

$\sin 45^\circ = \frac{1}{\sqrt{2}}$

$\sin 30^\circ = \frac{1}{2}$

Final Calculation:

Hence, the refractive index is:

$ \mu = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} $

Therefore, the refractive index of the liquid is $\sqrt{2}$.

Given:

Refractive index of the lens, $ \mu_l = 1.5 $

Refractive index of air, $ \mu_a = 1 $

Focal length of the lens in air, $ f = 20 \, \text{cm} $

Using the lens maker's formula:

$ \frac{1}{f} = \left(\frac{\mu_1}{\mu_a} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Substituting the known values:

$ \frac{1}{20} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

$ \frac{1}{10} = \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Now, when the lens is immersed in a liquid with refractive index $ \mu_{\text{liquid}} $, it behaves as a concave lens:

Power of the lens in the liquid, $ P = -1 \, \text{D} $

Calculating focal length in the liquid:

$ f = \frac{100}{P} = \frac{100}{-1} = -100 \, \text{cm} $

Applying the lens maker's formula for the lens immersed in the liquid:

$ \frac{-1}{100} = \left(\frac{1.5 - \mu_{\text{liquid}}}{\mu_{\text{liquid}}}\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Using the earlier result:

$ \frac{-1}{100} = \left(\frac{1.5}{\mu_{\text{liquid}}} - 1\right) \times \frac{1}{10} $

Solving for $ \mu_{\text{liquid}} $:

$ \frac{1.5}{\mu_{\text{liquid}}} = 1 - \frac{1}{10} $

$ \mu_{\text{liquid}} = \frac{1.5 \times 10}{9} $

$ \mu_{\text{liquid}} = \frac{5}{3} $

Focal length of objective $=2 \mathrm{~cm}$

Focal length of eyepiece eyepiece $=3 \mathrm{~cm}$ Distance between objective and eyepiece $=15 \mathrm{~cm}$

Final Image $=$ At infinity If the image is formed at infinity, then object should kept a focus [ $f=3 \mathrm{~cm}$ ] Hence distance of image formed by objective lens

$ 15-3=12 \mathrm{~cm} $

Using lens formula

$ \begin{aligned} & \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{2}=\frac{1}{12}-\frac{1}{x} \\ & x=-2.4 \mathrm{~cm} \end{aligned} $

Hence, option (a) is the correct option.

Given,

Convex lens (1) focal length $f_1=20 \mathrm{~cm}$

Convex lens (2) focal length $f_2=30 \mathrm{~cm}$

The focal length is positive for convex lens.

Focal length of the combination (when combined co-axially)

$ \begin{aligned} & \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{20}+\frac{1}{30} \\ & \frac{1}{f}=\frac{3+2}{60}=\frac{1}{12} \mathrm{~cm} \\ & f=12 \mathrm{~cm} \end{aligned} $

So, focal length of combined lens is 12 cm .

When upper half of the lens is painted black, then focal length of the lens will not change, hence the image will form on the same position. But, now the amount of the light forming image will be decreased, therefore, the intensity of image will decrease.

$ \begin{aligned} &\text { According to Snell's law, }\\ &\begin{aligned} \mu & =\frac{\sin i}{\sin r} \\ & =\frac{\sin 2 r}{\sin r} \quad(\because i=2 r) \\ & =\frac{2 \sin r \cos r}{\sin r} \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \mu=2 \cos r \Rightarrow \cos r=\frac{\mu}{2} \\ \Rightarrow & r=\cos ^{-1}\left(\frac{\mu}{2}\right) \\ \therefore & i=2 r=2 \cos ^{-1}\left(\frac{\mu}{2}\right) \end{array} $

Given,

Angle of incidence $=i$

Angle of emergence $=e$

Angle of prism $=A$

Let angle of deviation $=\delta$

$ \begin{aligned} & P Q=\text { incident ray } \\ & R S=\text { emergent ray } \end{aligned} $

Angle of deviation = the angle between the direction of incident ray and the emergent ray.

$ \begin{gathered} \delta=\angle U T S=\angle T Q R+\angle T R Q[\text { From } \triangle T Q R] \\ \delta=(\angle T Q V-\angle R Q V)+(\angle T R V-\angle Q R V) \\ \delta=(i-r)+(e-r)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{gathered} $

Now sum of angle of quadilateral $A Q V R=360^{\circ}$

$ \angle A Q V=\angle A R V=90^{\circ} $

$So, \angle A+\angle Q V R=180^{\circ} $

From $\triangle Q R V$

$ \begin{aligned} & r+r^{\prime}+\angle Q V R=180^{\circ} \\ & r+r^{\prime}=A \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eq. (i)

$ \delta=i+e-(r+r) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

Put the value of ( $r+r$ ) in Eq. (iii)

$ \delta=i+e-A $

Given,

Image distance, $v=-32 \mathrm{~cm}$

Object (sun) distance, $u=\infty$

Tank is filled upto $=20 \mathrm{~cm}$

Refractive index of water $=\frac{4}{3}$

Using mirror formula,

$ \begin{aligned} & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{-1}{32}+\frac{1}{\infty} \\ & f=-32 \mathrm{~cm} \end{aligned} $

When water is filled upto a height ( $h=20 \mathrm{~cm}$ ), the image formed by mirror will acts as virtual object for water surface.

The image formed by virtual object.

$ \begin{aligned} \frac{\mu_{\text {water }}}{\mu_{\text {air }}} & =\frac{\text { Actual height }}{\text { Apparent height }} \\ \frac{4}{3} & =\frac{X O}{X I} \Rightarrow X O=12 \mathrm{~cm} \\ \frac{4}{3} & =\frac{12}{X I}=9 \mathrm{~cm} \end{aligned} $

Sunlight will focus at 9 cm above water level.