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} $

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 $24^{\circ}$.

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.

Given, focal length of convex lens,

$ f=25 \mathrm{~cm} $

Refractive index, $\mu_g=1.5$

By using lens Maker's formula,

$ \frac{1}{f}=\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When cover lens is immersed in water, then focal length $\left(f_w\right)$ is given as

$ \frac{1}{f_w}=\mu\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Dividing Eq. (i) by Eq. (ii), we have

$ \begin{aligned} & \frac{\frac{1}{f}}{\frac{1}{f_w}}=\frac{\left(\mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}{\left(\frac{\mu_g}{\mu_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)} \\ \Rightarrow \quad & \frac{f_w}{f}=\frac{\left(\mu_g-1\right) \mu_w}{\left(\mu_g-\mu_w\right)}=\frac{(1.5-1) \frac{4}{3}}{\left(1.5-\frac{4}{3}\right)}=4 \\ \Rightarrow \quad & f_w=4 f=4 \times 25=100 \mathrm{~cm} \end{aligned} $

∴ Change in focal length

$ =f_w-f=100-25=75 \mathrm{~cm} $

Given, focal length of convex lens,

$ f=\frac{20}{3} \mathrm{~cm} $

Radii of curvature of double convex lens,

$ R_1=5 \mathrm{~cm}, R_2=-10 \mathrm{~cm} $

By lens Maker's formula,

$ \begin{aligned} & & \frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow & & \frac{1}{20} & =(\mu-1)\left(\frac{1}{5}-\frac{1}{-10}\right) \\ \Rightarrow & & \frac{3}{20} & =(\mu-1)\left(\frac{1}{5}+\frac{1}{10}\right) \\ \Rightarrow & & \frac{3}{20} & =(\mu-1)\left(\frac{3}{10}\right) \\ \Rightarrow & & \frac{1}{2} & =\mu-1 \\ \Rightarrow & & \mu & =1+\frac{1}{2} \\ & & & =\frac{3}{2}=1.5 \end{aligned} $

$ \begin{aligned} &\text { Use the formula, }\\ &\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R} \end{aligned} $

Given, $n_1=2, n_2=\sqrt{3}$

If $i_c$ be the critical angle, then

$ \begin{array}{r} \quad \sin i_c=\frac{1}{2} n_1=\frac{1}{\frac{n_1}{n_2}}=\frac{n_2}{n_1}=\frac{\sqrt{3}}{2} \\ \Rightarrow \quad \sin i_c=\sin 60^{\circ} \Rightarrow i_c=60^{\circ} \end{array} $

By using lens formula,

$ \begin{aligned} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\ & =\frac{1}{20}-\frac{1}{(-5)} \quad\binom{\because v=20 \mathrm{~cm}}{u=-5 \mathrm{~cm}} \\ & =\frac{1+4}{20}=\frac{1}{4} \end{aligned} $

When the glass plate of thickness $t(1.4 \mathrm{~cm})$ is inserted between the lens and the screen, then the shift in position of image

$ \begin{aligned} & =t\left(1-\frac{1}{\mu}\right) \\ & =1.4\left(1-\frac{1}{7 / 5}\right) \quad\left(\because \mu=\frac{7}{5}\right) \\ & =1.4\left(\frac{2}{7}\right)=0.4 \mathrm{~cm} \end{aligned} $

Thus, the lens forms the image at a distance of $20-0.4=19.6 \mathrm{~cm}$ from itself.

Hence, using again lens formula,

$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow & \frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{19.6}-\frac{1}{4}=\frac{3.9}{19.6}=\frac{39}{196} \\ \Rightarrow & u=\frac{196}{39}=5.02 \mathrm{~cm} \end{aligned} $

Refractive index, $\mu=1.5$

Since, one side of given lens is flat and other side is convex with a radius of curvature $R$.

$ \therefore R_1=R \text { and } R_2=\infty $

Position of object, $u=-60 \mathrm{~m}$

Position of image, $v=120 \mathrm{~cm}$

$ \begin{aligned} &\text { Using lens Maker's formula, we get }\\ &\begin{array}{ll} & \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow & \frac{1}{v}-\frac{1}{u}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\ & \quad\left(\because \frac{1}{f}=\frac{1}{v}-\frac{1}{u}\right) \\ \Rightarrow & \frac{1}{120}-\frac{1}{-60}=(1.5-1) \cdot \frac{1}{R} \\ \Rightarrow & \frac{3}{120}=\frac{0.5}{R} \Rightarrow \frac{1}{40}=\frac{0.5}{R} \\ \Rightarrow & R=40 \times 0.5=20 \mathrm{~cm} \end{array} \end{aligned} $

Let $\delta_m$ be the angle of minimum deviation.

As we know that,

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

Given, $\quad \mu=\sqrt{2}$ and $A=\delta_m$

$ \begin{array}{ll} \Rightarrow & \sqrt{2}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \frac{A}{2}}=\frac{\sin A}{\sin \frac{A}{2}} \\ \Rightarrow & \sqrt{2}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \\ & \quad \because \sin 2 A=2 \sin A \cos A] \\ & \quad \sqrt{2}=2 \cos \frac{A}{2} \Rightarrow \cos \frac{A}{2}=\frac{1}{\sqrt{2}} \\ \Rightarrow & \quad \frac{A}{2}=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right) \Rightarrow \frac{A}{2}=45^{\circ} \\ \Rightarrow & A=90^{\circ} \end{array} $

For dispersion without deviation through a prism combination,

$ \begin{aligned} \delta_{m_1} & =\delta_{m_2} \Rightarrow A_1\left(\mu_1-1\right)=A_2\left(\mu_2-1\right) \\ \Rightarrow A_2 & =A_1 \frac{\left(\mu_1-1\right)}{\left(\mu_2-1\right)} \end{aligned} $

Here, $A_1=9^{\circ}, \mu_1=1.4$ and $\mu_2=1.6$

$ \begin{aligned} \therefore \quad A & =A_2=9 \times \frac{(1.4-1)}{(1.6-1)} \\ & =9 \times \frac{0.4}{0.6}=6^{\circ} \end{aligned} $

The given situation is shown in the following figure.

$A B \rightarrow$ object, $A^{\prime} B^{\prime} \rightarrow$ image

$ A B=l, A^{\prime} B^{\prime}=l^{\prime}, u=-X, $

Hence, by mirror formula,

$ \begin{align*} & \frac{1}{-F} =\frac{1}{v}+\frac{1}{(-X)} \quad\left[\because \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\right] \\ \Rightarrow \frac{1}{v} & =\frac{1}{x}-\frac{1}{F} \Rightarrow \frac{1}{v}=\frac{F-X}{F X} \\ \Rightarrow & v =\frac{F X}{F-x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{align*} $

Longitudinal magnification for small object is given as

$ \begin{aligned} M & =\frac{l^{\prime}}{l}=-\frac{v^2}{u^2}=-\frac{\left(\frac{F X}{F-X}\right)^2}{(-X)^2} \\ & =-\frac{f^2 X^2}{(F-X)^2 X^2}=-\left(\frac{F}{F-X}\right)^2 \end{aligned} $

Magnitude of longitudinal magnification, $\frac{l^{\prime}}{l}=\left(\frac{F}{F-X}\right)^2$

The given situation is shown in the following figure.

For the ray incident at face $A C$ to be total internally reflected, if Angle of incidence $=$ Critical angle

$ \begin{aligned} & \Rightarrow \quad i=i_c \Rightarrow \sin (i)=\sin \left(i_c\right) \\ & \Rightarrow \quad \sin (90-\theta)=\frac{\mu_{\text {rarer }}}{\mu_{\text {denser }}} \\ & \Rightarrow \quad \cos (\theta)=\frac{\mu_l}{\mu_g} \Rightarrow \cos (\theta)=\frac{1.2}{1.5} \\ & \Rightarrow \quad \cos \theta=\frac{4}{5} \Rightarrow \cos \theta=0.8 \\ & \Rightarrow \quad \theta=\cos ^{-1}(0.8) \end{aligned} $

Magnification of a telescope for normal setting is

$ m=\frac{f_0}{f_e}=\frac{100}{5}=20 $

The given situation is shown below

For convex lens, $u=-8 \mathrm{~cm}, f=4 \mathrm{~cm}$

$ \begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{4}+\frac{1}{-8} \Rightarrow \frac{1}{v}=\frac{1}{8} \\ & \Rightarrow v=8 \mathrm{~cm} \end{aligned} $

For concave lens, image of convex lens acts like a virtual object.

$ \begin{aligned}So,\,\, & u=+2 \mathrm{~cm}, f=-4 \mathrm{~cm} \\ & \frac{1}{v}=\frac{1}{-4}+\frac{1}{2}=\frac{1}{4} \mathrm{~cm} \Rightarrow v=4 \mathrm{~cm} \end{aligned} $

So, distance from final image and object

$ =(8+6+4) \mathrm{cm}=18 \mathrm{~cm} $

The given situation is shown in the following figure

Given, $\quad v=+f, f=+f$

By $\quad \frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

We have, $\frac{1}{f}-\frac{1}{f}=\frac{1}{u} \Rightarrow \frac{1}{u}=0$

$ \text { ⇒ } \quad u=\infty $

So, object is at $\infty$ distance on right side of lens.