Ray Optics

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.

Using lens formula,

Given, $u_2+v_2=2.4 \mathrm{~m} \quad \text{.... (i)}$

Also, magnification for lens $(1)$ is 2.

$2=\frac{v_2+d}{u_2-d} \quad \text{... (ii)}$

From Eqs. (i) and (ii),

$u_2=0.8+d, v_2=1.6-d$

From lens formula,

$\frac{1}{v_2+d}+\frac{1}{u_2-d}=\frac{1}{f}$ (For lens (1))

$\frac{1}{v_2}+\frac{1}{u_2}=\frac{1}{f}$ (For lens (2))

$\begin{aligned} & \therefore \quad \frac{1}{v_2+d}+\frac{1}{u_2-d}=\frac{1}{v_2}+\frac{1}{u_2} \\ & \Rightarrow \quad \frac{1}{1.6-d+d}+\frac{1}{0.8+d-d}=\frac{1}{1.6-d}+\frac{1}{0.8+d} \\ & \Rightarrow \quad \frac{1}{1.6}+\frac{1}{0.8}=\frac{(0.8+d)+(1.6-d)}{(1.6-d)(0.8+d)} \end{aligned}$

$\begin{aligned} \Rightarrow \quad \frac{1+2}{1.6}=\frac{24}{(1.6-\mathrm{d})(0.8-d)} \\ \Rightarrow \quad \frac{3}{1.6}=\frac{2.4}{(1.6-d)(0.8-d)} \\ \Rightarrow \quad-d^2+0.8 d+1.28=1.28 \\ d^2-0.8 d=0 \end{aligned}$

$\begin{aligned} & d(d-0.8)=0 \\ \therefore \quad & d=0.8 \mathrm{~m} \\ \therefore \quad & \frac{1}{1.6-0.8}+\frac{1}{0.8+0.8}=\frac{1}{f} \\ \Rightarrow \quad & \frac{1}{0.53} \text { or } f=0.53 \mathrm{~m} \end{aligned}$

Let's denote the actual depth of the vessel as $d$. Given that the vessel is half-filled with one liquid and the other half with the second liquid, each liquid fills a depth of $\frac{d}{2}$.

Let the refractive index of the first liquid be $\mu$ and the refractive index of the second liquid be $2 \mu$.

The formula to find the apparent depth due to a single layer of liquid is:

$\text{Apparent Depth} = \frac{\text{Actual Depth}}{\text{Refractive Index}}$

First, we find the apparent depth for each segment of the vessel:

For the layer with refractive index $\mu$:

$\text{Apparent Depth}_1 = \frac{\frac{d}{2}}{\mu}$

For the layer with refractive index $2 \mu$:

$\text{Apparent Depth}_2 = \frac{\frac{d}{2}}{2 \mu}$

The total apparent depth is the sum of the apparent depths of both layers:

$\text{Total Apparent Depth} = \text{Apparent Depth}_1 + \text{Apparent Depth}_2$

$\text{Total Apparent Depth} = \frac{\frac{d}{2}}{\mu} + \frac{\frac{d}{2}}{2 \mu}$

$\text{Total Apparent Depth} = \frac{d}{2 \mu} + \frac{d}{4 \mu}$

$\text{Total Apparent Depth} = \frac{2d + d}{4 \mu}$

$\text{Total Apparent Depth} = \frac{3d}{4 \mu}$

We are given that the apparent depth of the vessel is $50 \%$ of the actual depth. Therefore:

$\frac{3d}{4 \mu} = \frac{d}{2}$

We solve for $\mu$ by isolating it on one side of the equation. Multiply both sides by $4 \mu$:

$3d = 2d \mu$

Divide both sides by $2d$:

$\mu = \frac{3}{2}$

$\mu = 1.5$

So, the correct option is:

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} $

For water, real depth $=12 \mathrm{~cm}$

Apparent depth $=9 \mathrm{~cm}$

Since, refractive index, $\mu_w=\frac{\text { Real depth }}{\text { Apparent depth }}$

$\Rightarrow \quad \mu_{\mathrm{w}}=\frac{12}{9}$

when water is replaced by liquid of refractive index $\mu=1.5$, we have

New apparent depth $=\frac{\text { real depth }}{\mu_l}=\frac{12}{1.5}=8 \mathrm{~cm}$

Hence, required shifted distance $=9-8=1 \mathrm{~cm}$

For double convex lens,

$R_1=4 \mathrm{~cm}, R_2=-8 \mathrm{~cm}$

Refractive index, $\mu=1.5$

By lens maker's formula, focal length is given by

$\begin{aligned} \frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & =(1.5-1)\left(\frac{1}{4}-\frac{1}{-8}\right)=0.5\left(\frac{1}{4}+\frac{1}{8}\right) \\ & =0.5\left(\frac{3}{8}\right)=\frac{3}{16} \\ \Rightarrow \quad f & =\frac{16}{3} \mathrm{~cm} \\ & =5.33 \mathrm{~cm} \end{aligned}$

From the given situation abṣolute refractive index of denser medium, ${ }^a \mu_d=2$

Absolute refractive index of rarer medium, ${ }^a \mu_r=1$

$\therefore$ Refractive index of denser medium with respect to rarer medium

${ }^r \mu_d=\frac{{ }^a \mu_d}{{ }^a \mu_r}=\frac{2}{1}=2 \Rightarrow{ }^r \mu_d=2$

If $i_c$ be the critical angle, then

$\begin{aligned} \sin i_c & =\frac{1}{{ }_r^r \mu_d}=\frac{1}{2}=\sin 30^{\circ} \\ \Rightarrow \quad i_c & =30^{\circ} \end{aligned}$

As, we know, ${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$

$ \Rightarrow {{1.5} \over v} - {1 \over {( - 20)}} = {{1.5 - 1} \over { + 40}}$

$ \Rightarrow v = - 40$ cm

Negative sign shows that image is obtained on the same side of object i.e., towards left.

To find the ratio of the dispersive power of the convergent and divergent lenses in an achromatic doublet, we need to use some fundamental formulas from optics. Let’s denote the power of the converging lens by P₁ = +6 D and the power of the diverging lens by P₂.

Given that the total power of the achromatic doublet is P = +5 D, we can write the equation for the total power as:

$ P = P_1 + P_2 $

Substituting the given values:

$ 5 = 6 + P_2 $

Solving for P₂:

$ P_2 = 5 - 6 = -1 \, \text{D} $

Next, according to the condition for achromatic lenses, the product of the power of the lenses and their dispersive powers must be equal and of opposite sign. Let \bomega be the dispersive power of the converging lens, and \bomega’ be the dispersive power of the diverging lens:

$ P_1 \cdot \omega_1 = -P_2 \cdot \omega_2 $

Substituting the power values we found:

$ 6 \cdot \omega_1 = -(-1) \cdot \omega_2 $

$ 6 \cdot \omega_1 = \omega_2 $

From this equation, we can find the ratio of the dispersive powers:

$ \frac{\omega_1}{\omega_2} = \frac{1}{6} $

Thus, the ratio of the dispersive power of the convergent lens to the divergent lens is 1 : 6, which corresponds to Option C.

So the correct answer is:

Option C

1 : 6

By using lens Maker formula,

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

where,

$\begin{aligned} & f=\text { focal length, } \\ & \mu=\text { refractive index, } \end{aligned}$

$R_1$ and $R_2=$ radii of curvature.

If light changes from red to blue, refractive index of light also changes. Hence, focal length will also change.

Hence, Assertion and Reason both are false.

Given, wavelength of light $A, \lambda_A=300 \mathrm{~nm}$

$=300 \times 10^{-9} \mathrm{~m}$

Frequency of light in $A, f_A=5 \times 10^{14} \mathrm{~Hz}$

Ratio of speed of light $\frac{v_A}{v_B}=\frac{4}{5}$

Let $\mu_A, \mu_{\text {air }}$ and $\mu_B$ are refractive indexes of medium $A$, air and $B$ respectively.

$\begin{aligned} & \because \text { Absolute refractive index }=\frac{\mu_A}{\mu_{\text {air }}} \\ & =\frac{\text { Speed of light in air }(c)}{\text { Speed of light in medium } A\left(v_A\right)} \\ & \Rightarrow \quad \mu_A=\frac{3 \times 10^8}{f_A \lambda_A} \\ & \Rightarrow \quad \mu_A=\frac{3 \times 10^8}{5 \times 10^{14} \times 300 \times 10^{-9}}=0.2 \times 10^1=2 \\ \end{aligned}$

Now, $\frac{\mu_B}{\mu_A}=\frac{v_A}{v_B}$

$\Rightarrow \quad \mu_B=\mu_A\left(\frac{v_A}{v_B}\right)=2\left(\frac{4}{5}\right)=\frac{8}{5}=1.6 $

In fibre optic communication, signals are transmitted through an optical fibre. The property of light used in transmission through optical fibre cables is total internal reflection.

Let N be the maximum number of reflection that ray undergoes.

Suppose it covers x distance in one reflection, then

$Nx = 2\sqrt 3 $

$ \Rightarrow N = {{2\sqrt 3 } \over x} = {{2\sqrt 3 } \over {0.2\,\tan 30^\circ }}$ [$\because$ $x = 0.2\,\tan 30^\circ $]

$N = {6 \over {0.2}} = 30$

Number of reflections ray undergoes before it emerges excluding the first one are N $-$ 1

$ = 30 - 1 = 29$

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.

The focal length of a convex lens of refractive index $\mu$_g in air,

${1 \over {{f_{air}}}} = ({\mu _g} - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ ...... (i)

where, R₁, R₂ = radii of curvatures of its first and second surface.

When the lens is immersed in a liquid of refractive index $\mu$_I, then refractive index of material of lens (glass) w.r.t. liquid is,

${}_I{\mu _g} = {{{\mu _g}} \over {{\mu _I}}}$ ..... (ii)

Now, focal length of lens in liquid

${1 \over {f'}} = ({}_I{\mu _g} - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ ...... (iii)

Dividing Eq. (i) by Eq. (iii), we get

${{f'} \over {{f_{air}}}} = {{{\mu _g} - 1} \over {({}_I{\mu _g} - 1)}} = {{{\mu _g} - 1} \over {\left( {{{{\mu _g}} \over {{\mu _I}}} - 1} \right)}}$

Given, f' = 0.5 m, f_air = 0.2 m, $\mu$_g = 1.5

Substituting all these values in above equation, we get

$ - {{0.5} \over {0.2}} = {{1.5 - 1} \over {\left( {{{1.5} \over {{\mu _1}}} - 1} \right)}} = {{0.5} \over {\left( {{{1.5} \over {{\mu _1}}} - 1} \right)}}$

$ \Rightarrow {{1.5} \over {{\mu _I}}} - 1 = - 0.2 \Rightarrow {{1.5} \over {{\mu _I}}} - 1 - 0.2 = 0.8$

$ \Rightarrow {\mu _I} = {{15} \over 8}$

$\therefore$ Refractive index of liquid $ = {{15} \over 8}$

Power of first spherical lens, P₁ = $-$4 D

The power of this lens is negative, hence it is diverging lens.

Focal length of first spherical lens,

${f_1} = {1 \over {{P_1}}} = {1 \over {( - 4D)}} = - 0.25$ m = $-$25 cm

Power of second spherical lens, P₂ = 5 D

Power of this lens is positive, hence it is converging lens.

Focal length of second spherical lens,

${f_2} = {1 \over {{P_2}}} = {1 \over {5D}} = 0.2$ m = 20 cm

The given situation is shown in the following figure.

Now, the image I₁ will be considered as object for converging lens.

For converging lens, distance of object,

u = $-$(25 + 15) = $-$40 cm and f₂ = 20 cm

$\therefore$ Using lens formula,

${1 \over {{f_2}}} = {1 \over v} - {1 \over u} \Rightarrow {1 \over {20}} - {1 \over v} - {1 \over { - 40}}$

$\Rightarrow$ v = 40 cm

Since v is positive, so image formed will be real and will be formed at a distance of 40 cm from the lens of power 5 D (converging lens).