Ray Optics

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, 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} $

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

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

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 $