Ray Optics

As focal length of convex lens is 10 cm , so image of an object placed 20 cm is formed 20 cm beyond lens.

These rays are then diverged by concave lens and so a virtual image $I$ is finally formed between both lenses.

Velocity of light waves in material is

$ v=\nu \lambda \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Refractive index of material is

$ \mu=\frac{c}{v} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

where, $c$ is speed of light in vacuum or air.

or   $ \mu=\frac{c}{v \lambda} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} Given,\,\, \nu & =2 \times 10^{14} \mathrm{~Hz} \\ \lambda & =5000 \mathop {\rm{A}}\limits^{\rm{o}} =5000 \times 10^{-10} \mathrm{~m} \\ c & =3 \times 10^8 \mathrm{~ms}^{-1} \\ \mu & =\frac{3 \times 10^8}{2 \times 10^{14} \times 5 \times 10^{-7}}=3 \end{aligned} \end{aligned} $

Power of an equiconvex lens is given as $ P $ diopters, which relates to its focal length $ f $ by the equation $ P = \frac{1}{f} $.

When this equiconvex lens is divided into two symmetrical halves using a plane that includes the principal axis, each half retains the same focal length $ f $. As a result, the power of each half-lens remains the same as the original lens, maintaining the relation:

$ P' = \frac{1}{f} \quad \Rightarrow \quad P' = P $

Therefore, the ratio of the power of each half to the original lens power is:

$ \frac{P'}{P} = 1 $

Thus, the ratio of the power of the two halves is $ 1:1 $.

The magnifying power of a telescope is given as 9, with the distance between the objective and eyepiece being 20 cm when adjusted for parallel rays. We need to find the ratio of the focal length of the objective lens to the focal length of the eyepiece, denoted as $ m $.

To determine this ratio:

Magnifying Power Formula:

The magnifying power of a telescope $ m $ is defined by the formula:

$ m = \frac{f_0}{f_e} $

where $ f_0 $ is the focal length of the objective lens and $ f_e $ is the focal length of the eyepiece lens.

Given Values:

$ m = 9 \Rightarrow \frac{f_0}{f_e} = 9 $

Thus, we can express the focal length of the objective lens as:

$ f_0 = 9f_e $

Distance Between Lenses:

The total distance between the objective and eyepiece is given by:

$ f_0 + f_e = 20 \text{ cm} $

Substitute and Solve:

Substitute the expression for $ f_0 $ into the distance equation:

$ 9f_e + f_e = 20 $

$ 10f_e = 20 $

Solving this gives:

$ f_e = 2 \text{ cm} $

Finding $ f_0 $:

Using the relationship between $ f_0 $ and $ f_e $:

$ f_0 = 9 \times 2 = 18 \text{ cm} $

Ratio Calculation:

The ratio of the focal lengths is:

$ \frac{f_0}{f_e} = \frac{18}{2} = 9 $

Therefore, the value of $ m $ is 9.

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:

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

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$

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).