Geometrical Optics

$ \begin{aligned} &i=e\\ &\text { Equation of prism }\\ &\begin{aligned} & i+e=A+\delta \\ & 2 i-A=\delta \\ & 2(50)-60=\delta \\ & \delta=40^{\circ} \end{aligned} \end{aligned} $

$ F_2 \text { is the second principal focus ⇒ It is the virtual image position for object at infinity. } $

$F_1$ is the first principal focus ⇒ It is the virtual object position for which image is formed at infinity. The best appropriate answer is option (2), although it should be second principal focus.

$ \text { (i) When object } O \text { is viewed from medium } P \text {, } $

Apparent depth $h_1=\frac{h_{\text {real }}}{\mu}=\frac{2.5}{1.25}=2 \mathrm{~cm}$

(ii) When object $O$ is viewed from medium $R$, Apparent depth $h_2={ }_Q \mu_R \times 2.5$

$ \begin{aligned} & =2.5 \times \frac{1.5}{1.25}=3 \mathrm{~cm} \\ & \therefore\left|h_1-h_2\right|=1 \mathrm{~cm} \end{aligned} $

For convex lens $L_1: u=-30 \mathrm{~cm}$ and $f=+10 \mathrm{~cm}$

For concave lens $L_2: u^{\prime}=+12 \mathrm{~cm}$ and $f^{\prime}=-10 \mathrm{~cm}$ Using $\frac{1}{v^{\prime}}-\frac{1}{u^{\prime}}=\frac{1}{f^{\prime}} \Rightarrow \frac{1}{v^{\prime}}-\frac{1}{12}=\frac{-1}{10} \Rightarrow \frac{1}{v^{\prime}}=\frac{1}{12}-\frac{1}{10}=\frac{-1}{60}$

⇒ Final image formed at 60 cm to the left of concave lens.

Index correction is an adjustment which need to perform in the experiments related to optical bench while measuring the focal length of lenses using optical bench.

To determine the magnification of the microscope, we use the formula:

$ m = \frac{L}{f_o} \times \frac{D}{f_e} $

Where:

$ m $ is the total magnification.

$ L $ is the tube length of the microscope.

$ f_o $ is the focal length of the objective lens.

$ D $ is the distance of distinct vision of the eye.

$ f_e $ is the focal length of the eyepiece.

Given:

$ L = 40 $ cm (tube length)

$ f_o = 2 $ cm (focal length of the objective)

$ D = 25 $ cm (distance of distinct vision)

$ f_e = 4 $ cm (focal length of the eyepiece)

Substitute these values into the formula:

$ m = \frac{40}{2} \times \frac{25}{4} $

Calculate each part:

$ \frac{40}{2} = 20 $

$ \frac{25}{4} = 6.25 $

Multiply these results:

$ m = 20 \times 6.25 = 125 $

Thus, the magnification of the microscope is 125.

In a camera with a series combination of four similar thin convex lenses, the effective power and total magnification of this combination can be calculated as follows:

Effective Power: When lenses are in contact, their powers add up. Therefore, the effective power $ p_{\text{eff}} $ of the combination is the sum of the powers of each individual lens:

$ p_{\text{eff}} = p_1 + p_2 + p_3 + p_4 = 4p $

Total Magnification: The total magnification $ m_{\text{eff}} $ is the product of the magnifications of each lens in the series:

$ m_{\text{eff}} = m_1 \times m_2 \times m_3 \times m_4 = m^4 $

Thus, for a combination of four identical lenses, the effective power is four times the power of a single lens, and the effective magnification is the fourth power of the magnification of a single lens.

The correct answer is Option C: Both Statement I and Statement II are correct.

Here's why:

Statement I: Image formation needs regular reflection and/or refraction.

This statement is correct. Image formation relies on the interaction of light with surfaces. Here's how:

• Regular Reflection: This occurs when parallel rays of light strike a smooth surface and reflect back in a parallel manner. Mirrors are a prime example of regular reflection. This type of reflection allows for clear, focused images.


• Refraction: This occurs when light passes from one medium to another (like from air to water or air to glass). The change in speed causes the light to bend, resulting in image formation. This is how lenses in cameras, telescopes, and our own eyes work.

Statement II: The variety in colour of objects we see around us is due to the constituent colours of the light incident on them.

This statement is also correct. The colors we perceive are based on the interaction between light and objects:

• White Light: Sunlight or light from a bulb contains all the colors of the rainbow (the visible spectrum).


• Selective Absorption: Objects absorb certain colors of light and reflect others. For instance, a red apple absorbs most colors except red, which it reflects back to our eyes.


• Pigments: Many objects, especially those we paint or dye, contain pigments that absorb specific colors and reflect others.

In summary, both statements are essential to understanding how we see images and perceive colors.

In prism, $r_1+c=A$

$\begin{aligned} \quad & r_1=90^{\circ}-c \quad \text{.... (1)}\\ & \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$

$\Rightarrow$ Apply Snell's law, on incidence surface

$\begin{aligned} 1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right) \\ \frac{1}{2}=\mu \times \frac{\sqrt{\mu^2-1}}{\mu} \end{aligned}$

On squaring $\frac{1}{4}=\mu^2-1$

$\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}$

The magnifying power of a telescope when viewing distant objects can be calculated using the formula:

$ \text{Magnifying Power} (M) = \frac{\text{Focal Length of Objective Lens} (f_o)}{\text{Focal Length of Eyepiece Lens} (f_e)} $

Given in the problem:

• Focal Length of Objective Lens, $f_o = 140 \mathrm{~cm}$
• Focal Length of Eyepiece Lens, $f_e = 5.0 \mathrm{~cm}$

Substituting these values into the formula gives:

$ M = \frac{140 \mathrm{~cm}}{5.0 \mathrm{~cm}} = 28 $

Thus, the magnifying power of the telescope is 28. The correct answer is Option B.

Since lens is made of three materials so three $\mu$ and hence three images.

$\mu=\frac{C}{\mathrm{~V}}=\frac{\frac{1}{\sqrt{\mu_0 \in_0}}}{\frac{1}{\sqrt{1.5 \mu_0 \times 2 \epsilon_0}}}=\sqrt{3}$

$A = {r_1} + {r_2} = 6^\circ $

$\mu = 1.5{{\sin e} \over {\sin {r_2}}} = {{\sin e} \over {\sin 6}}$

$\sin e = 1.5 \times 6^\circ $

$e \simeq 9^\circ $

$\mathrm{x=2f}$

$\mathrm{f=x/2}$

The critical angle, denoted as $\theta_c$, is the angle of incidence beyond which light is totally internally reflected within a denser medium when it hits the boundary with a less dense medium. To find the critical angle for the medium in question, first, we need to understand the relationship between the speed of light in different media and their refractive indices.

Let's denote the speed of light in air as $V_1$ and in the denser medium as $V_2$. From the given information:

$V_1 = \frac{x}{t_1}$ (speed of light in air)

$V_2 = \frac{10x}{t_2}$ (speed of light in the denser medium)

The refractive index of a medium (n) is inversely proportional to the speed of light in that medium ($n = \frac{c}{V}$, where c is the speed of light in vacuum). Thus, the refractive index of the denser medium ($n_2$) relative to air ($n_1$) can be obtained by taking the ratio of the speed of light in air to that in the denser medium:

$\frac{n_2}{n_1} = \frac{V_1}{V_2}$

Substituting the expressions for $V_1$ and $V_2$:

$\frac{n_2}{n_1} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} = \frac{t_2}{10t_1}$

For the critical angle $\theta_c$, the light in the denser medium (index $n_2$) strikes the boundary with the less dense medium (index $n_1$) such that it refracts at 90 degrees (escapes along the boundary). Snell's Law at this boundary is:

$n_2 \sin \theta_c = n_1 $

Given that $n_1 = 1$ (approximating the refractive index of air), we simplify to:

$\sin \theta_c = \frac{1}{n_2}$

Substituting $n_2$ from the earlier relation and simplifying:

$\sin \theta_c = \frac{10t_1}{t_2}$

Hence, the critical angle $\theta_c$ is given by:

$\theta_c = \sin^{-1}\left(\frac{10 t_1}{t_2}\right)$

Use $\frac{1}{\mathrm{f}}=[\mu-1]\left[\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right]$

$\frac{1}{\mathrm{f}_{1}}=[1.6-1]\left[\frac{1}{\infty}-\frac{1}{20}\right]=\frac{-3}{100}$

$\frac{1}{\mathrm{f}_{2}}=[1.5-1]\left[\frac{1}{20}+\frac{1}{20}\right]=\frac{1}{20}$

$\frac{1}{\mathrm{f}_{3}}=\frac{-3}{100}$

$\frac{1}{f_{e q}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}$

$\frac{1}{f_{e q}}=-\frac{3}{100}+\frac{1}{20}-\frac{3}{100}=\frac{-1}{100}$

$\frac{1}{f_{e q}}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

$\frac{1}{f_{e q}}=\frac{1}{f}-\frac{1}{f}$

$\mathrm{f}_{\mathrm{eq}}=\infty$

Primary rainbow is result of three-step process, that is, refraction, internal reflection and refraction.

Secondary rainbow is result of four step process, that is, refraction, internal reflection, internal reflection and refraction.

Secondary rainbow appears above the primary rainbow.

Given f₀ = 20 m = 2000 cm, f_e = 2 cm

$\bullet$ Distance between objective and eye piece

l = f₀ + f_e

= 20 m + 2 cm = 20.02 cm

$\bullet$ Magnification of telescope

$m = - {{{f_0}} \over {{f_e}}} = - {{2000\,cm} \over {2\,cm}} = - 1000$

$\bullet$ Image formed by telescope is inverted.

$\bullet$ Aperture of eye piece is smaller than that of objective.

So, statements (a), (b) and (c) are correct.

Power of lens is given by

$P = {1 \over {f(m)}}$

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

${1 \over f} = \left\{ {{3 \over 2} - 1} \right\}\left( {{1 \over {20}} + {1 \over {20}}} \right)$

$f = 20$ cm

$P = {1 \over {20 \times {{10}^{ - 2}}}}$

$ = 5$ D

Given i = 60$^\circ$ and $\mu$ = $\sqrt3$

$\Rightarrow$ Here, angle of incidence $\Rightarrow$ i = tan^$-$1 ($\mu$)

Hence, reflected and refracted rays would be perpendicular to each other.

${\mu _A} = {{3 \times {{10}^8}} \over {1.5 \times {{10}^8}}} = 2$

${\mu _B} = {{3 \times {{10}^8}} \over {2 \times {{10}^8}}} = 1.5$

For TIR, ray of light should travel from denser to rarer medium

${\mu _A}\sin {\theta _C} = {\mu _B}\sin 90^\circ $

$2\sin {\theta _C} = 1.5\sin 90^\circ $

$\sin {\theta _C} = 0.75$

${\theta _C} = {\sin ^{ - 1}}(0.75)$

We know, ${1 \over f} = (\mu - 1)\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right)$

For the lens on the left,

${1 \over {{f_1}}} = (1.5 - 1)\left( {{1 \over \infty } - {1 \over { - 20}}} \right) = {1 \over {40}}$

For the lensh on the right,

${1 \over {{f_2}}} = (1.5 - 1)\left( {{1 \over { + 20}} - {1 \over \infty }} \right) = {1 \over {40}}$

For the oil in the intervening space,

${1 \over {{f_3}}} = (1.7 - 1)\left( {{1 \over { - 20}} - {1 \over {20}}} \right) = - {7 \over {100}}$

Hence, if the focal length of the combination is F, then

${1 \over F} = {1 \over {{f_1}}} + {1 \over {{f_2}}} + {1 \over {{f_3}}} = {1 \over {40}} + {1 \over {40}} - {7 \over {100}} = - {1 \over {50}}$

i.e., F = $-$50 cm