Geometrical Optics

Parallel rays are focussed by convex lens and viceversa.

$\begin{aligned} \therefore \quad L_1 L_2 & =f_1+f_2 \\ & =10+15=25 \mathrm{~cm} \end{aligned}$

Let's analyze each statement carefully:

Statement (I): When an object is placed at the centre of curvature of a concave lens, image is formed at the centre of curvature of the lens on the other side.

To understand this statement, let's recall the nature of images formed by concave lenses. A concave lens (diverging lens) always forms images that are virtual, erect, and smaller than the object, regardless of the object's position. The term "centre of curvature" is typically associated with mirrors rather than lenses. Therefore, this statement is incorrect because a concave lens does not form a real image that could be said to be at the centre of curvature on the other side.

Statement (II): Concave lens always forms a virtual and erect image.

Now, considering the second statement, a concave lens (diverging lens) indeed always forms virtual, erect, and diminished images. This is a fundamental property of concave lenses.

Given this analysis:

The correct answer is Option A: Statement I is false but Statement II is true.

$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ & \frac{1}{v}-\frac{1}{-30}=\frac{1}{10} \\ & \frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{3-1}{30}=\frac{1}{15} \\ & v=15 \end{aligned}$

$\Rightarrow 30 \mathrm{~cm} \text { right of third }$

The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $90^{\circ}$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:

$ n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) $

Here, $n_1$ is the refractive index of the first medium, $n_2$ is the refractive index of the second medium, and $\theta_c$ is the critical angle. Since $\sin(90^{\circ}) = 1$, the equation simplifies to:

$ n_1 \sin(\theta_c) = n_2 $

Given that the critical angle $\theta_c$ is $45^{\circ}$, we have:

$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $

Substituting this value in the simplified Snell's Law equation, we get:

$ n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2 $

Therefore,

$ n_1 = n_2 \cdot \frac{2}{\sqrt{2}} $

$ n_1 = n_2 \cdot \sqrt{2} $

Hence, the ratio of the refractive indices of the first and second media is:

$ n_1 : n_2 = \sqrt{2} : 1 $

The correct answer is therefore:

Option D

$\sqrt{2}: 1$

To find the refractive index of the material of a thin convex lens, we can make use of the Lensmaker's Formula. The Lensmaker's formula is given by:

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

where

• $f$ is the focal length of the lens,
• $\mu$ is the refractive index of the material of the lens,
• $R_1$ is the radius of curvature of the first surface (convex side, positive),
• $R_2$ is the radius of curvature of the second surface (concave side, negative for convex lens).

Given in the question, the radii of curvature are $15 \, \mathrm{cm}$ and $30 \, \mathrm{cm}$ respectively, and the focal length $f = 20\, \mathrm{cm}$.

It's important to pay attention to the signs of the radii of curvature according to the lens maker's convention. For convex lenses, $R_1$ is positive and $R_2$ is negative; however, since the problem doesn't specify which curvature corresponds to which side in relation to the direction of light travel, we'll assume the light travels from left to right: thus, $R_1 = +15 \, \mathrm{cm}$ and $R_2 = -30 \, \mathrm{cm}$.

Substituting the given values into the Lensmaker's Formula, we get:

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} - \frac{1}{-30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{15} + \frac{1}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{2 + 1}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{3}{30} \right) $

$\frac{1}{20} = (\mu - 1) \left( \frac{1}{10} \right) $

$\frac{1}{20} = \frac{\mu - 1}{10}$

Now, solve for $\mu$:

$\mu - 1 = \frac{10}{20} $

$\mu - 1 = 0.5 $

$\mu = 1.5$

Hence, the refractive index of the material of the lens is 1.5, which corresponds to Option B.

When we have a combination of identical lenses in contact, the effective power ($P_{\text{eff}}$) of the combination can be calculated as the sum of the powers of all the individual lenses. This is because the lenses are in direct contact, and their powers effectively add up.

Given that the effective power of a combination of 5 identical convex lenses is $25 \mathrm{D}$, we can use the formula for the effective power of the combination:

$P_{\text{eff}} = nP$

where:

• $P_{\text{eff}}$ is the effective power of the combination,
• $n$ is the number of lenses, and
• $P$ is the power of each individual lens.

Given $n = 5$ and $P_{\text{eff}} = 25 \mathrm{D}$, we can solve for $P$, the power of each lens:

$25 \mathrm{D} = 5P$

Dividing both sides by 5:

$P = \frac{25 \mathrm{D}}{5} = 5 \mathrm{D}$

The power of a lens ($P$) is related to its focal length ($f$) by the equation:

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

where $P$ is in diopters (D) and $f$ is in meters. Thus:

$5 \mathrm{D} = \frac{1}{f}$

Solving for $f$ gives:

$f = \frac{1}{5} \text{ meters} = \frac{1}{5} \times 100 \text{ cm} = 20 \text{ cm}$

Therefore, the focal length of each of the convex lenses is 20 cm. So, the correct answer is Option B: 20 cm.

First, let's understand the relationship between the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a convex lens, which is given by the lens formula:

Now, to find the error in the focal length ($\Delta f$) due to the errors in the measurements of $u$ and $v$ ($\Delta u$ and $\Delta v$, respectively), we have to differentiate the lens formula with respect to $u$ and $v$, keeping in mind the propagation of error.

By differentiating both sides of the lens formula with respect to $v$ and $u$, and also considering the negative reciprocal relation (given $f$ is a constant for a specific lens), we have:

Rearranging this equation to find $\Delta f$, we get:

Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts ($\Delta u$ and $\Delta v$) of the measuring scales for the position of the object ($u$) and for the position of the image ($v$), is:

Option C: $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$

$\begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\ & \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\ & \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta m}{2} \\ & \delta_m=\pi-2 A \end{aligned}$

$\begin{aligned} & \mathrm{m}=2=\frac{-v}{u} \\ & 2=\frac{-(15-u)}{-u} \\ & 2 u=15-u \\ & 3 u=15 \Rightarrow u=5 \mathrm{~cm} \\ & v=15-u=15-5=10 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ & =\frac{1}{10}+\frac{1}{(-5)}=\frac{1-2}{10}=\frac{-1}{10} \\ & f=-10 \mathrm{~cm} \end{aligned}$

Given $\mathrm{R}=30 \mathrm{~cm}$

$\mathrm{f}=\mathrm{R} / 2=+15 \mathrm{~cm}$

Magnification $(\mathrm{m})= \pm \frac{1}{2}$

For convex mirror, virtual image is formed for real object.

Therefore, $\mathrm{m}$ is +ve

$\begin{aligned} & \frac{1}{2}=\frac{f}{f-u} \\ & u=-15 \mathrm{~cm} \end{aligned}$

$\begin{aligned} & \mu_1=1.5 \\ & \mu_m=1.6 \\ & f_a=20 \mathrm{~cm} \\ & \text { As } \frac{f_m}{f_a}=\frac{\left(\mu_1-1\right) \mu_m}{\left(\mu_1-\mu_m\right)} \\ & \frac{f_m}{20}=\frac{(1.5-1) 1.6}{(1.5-1.6)} \\ & f_m=-160 \mathrm{~cm} \end{aligned}$

$\begin{aligned} & \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\ & \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\ & \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\ & \delta_{\min }=\pi-2 \mathrm{~A} \end{aligned}%$

Spherometer can be used to measure curvature of surface.

As amount of energy incident on cell is same so current will remain same.

To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is $90^{\circ}$. Snell's Law states:

$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$

where $n_1$ and $n_2$ are the indices of refraction for the denser and rarer media, respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

In this case, we have:

$n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}$

The critical angle is given as $45^{\circ}$, and the speed of light in the rarer medium is given as $3 \times 10^8 \mathrm{~m/s}$. We can find the index of refraction of the rarer medium using the formula:

$n = \frac{c}{v}$

where $c$ is the speed of light in a vacuum ($3 \times 10^8 \mathrm{~m/s}$), and $v$ is the speed of light in the medium.

For the rarer medium, we have:

$n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1$

Now we can rewrite Snell's Law as:

$n_1 \sin{45^{\circ}} = 1 \cdot 1$

$n_1 \sin{45^{\circ}} = 1$

$n_1 = \frac{1}{\sin{45^{\circ}}}$

Since $\sin{45^{\circ}} = \frac{1}{\sqrt{2}}$, we have:

$n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

Now, we can find the speed of light in the denser medium using the formula for the index of refraction:

$v_1 = \frac{c}{n_1}$

$v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}$

$v_1 = 2.12 \times 10^8 \mathrm{~m/s}$

So, the speed of light in the denser medium is $2.12 \times 10^8 \mathrm{~m/s}$.

Reflecting telescopes, also known as reflectors, use a set of mirrors instead of lenses to gather and focus light. The primary mirror (usually a concave mirror) gathers the incoming light and reflects it to a focus point. The secondary mirror is used to redirect this focused light out to where it can be conveniently observed.

In other words, the secondary mirror in a reflecting telescope is used to move the eyepiece outside the telescopic tube, where the image can be comfortably viewed. This is especially important for large telescopes, where the focus point may be inside the telescope tube and inaccessible without a secondary mirror.

Assertion A is correct. When light waves travel through different media with different indices of refraction, the speed of light changes, which in turn changes the phase of the light waves. This is because the phase of a wave is directly related to the distance it has traveled, which in this case is affected by the speed of light in the different media.

Reason R is also correct. The speed of light in a medium is determined by its refractive index. When light enters a medium with a different refractive index, its speed, and therefore its wavelength, changes. This is described by the equation v = c/n, where v is the speed of light in the medium, c is the speed of light in a vacuum, and n is the refractive index of the medium.

Moreover, R is indeed the correct explanation of A. The change in wavelength (and therefore phase) of the light waves in different media (as explained in R) is the reason why the phase difference changes when light waves travel through different media with the same thickness but different indices of refraction (as stated in A).

In this problem, you are asked to find the percentage error in the estimation of the focal length of a convex lens using a 2-meter long scale with a least count of 0.2 cm.

First, let's determine the object distance (u), image distance (v), and focal length (f) of the lens.

1. Object distance (u): It's the distance between the object pin and the convex lens. The object pin is at the 80 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the object distance is $u = 100 - 80 = 20~cm$.

2. Image distance (v): It's the distance between the image pin and the convex lens. The image pin is at the 180 cm mark, and the convex lens is at the 1 m (100 cm) mark, so the image distance is $v = 180 - 100 = 80~cm$.

3. Focal length (f): Using the lens formula, we can calculate the focal length:
   
   $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{5}{80}$ So, $f = \frac{80}{5} = 16~cm$.

Now, we will calculate the error in the focal length (df) using the given least count (0.2 cm). The error in the object distance and image distance will both be 0.2 cm.

1. Error in the focal length (df): We can use the formula for the error in the focal length:
   
   $\frac{df}{f^2} = \frac{0.2 \times 2}{6400} + \frac{0.2 \times 2}{400}$

Solving for df:

$df = \frac{16 \times 16 \times 0.2 \times 6800 \times 2}{6400 \times 400} = 0.136 \times 2$

1. Percentage error in the focal length: Finally, we will calculate the percentage error using the formula:
   
   $\frac{df}{f} = \frac{0.0085 \times 2}{1} = 1.70$

So, the percentage error in the estimation of the focal length is 1.70%.

When a light wave moves from one medium to another, its speed and wavelength may change, but its frequency remains constant because it is determined by the source of the light. This is because frequency depends on the oscillations of the source, which do not change when entering a different medium.

Snell's law, which describes the relationship between the angles of incidence and refraction, can be written as:

$ n_1 \sin(\theta_1) = n_2 \sin(\theta_2), $

where $n_1$ and $n_2$ are the refractive indices of the two media, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively.

The refractive index of a medium is also related to the speed of light in that medium:

$ n = \frac{c}{v}, $

where $c$ is the speed of light in a vacuum, and $v$ is the speed of light in the medium.

From the above relations, we see that as light enters a medium with a higher refractive index (and hence a lower speed), its wavelength decreases. The frequency remains the same.

Given that the angle of incidence is $45^\circ$ and the angle of refraction is $30^\circ$, the ratio of the refractive indices is:

$ \frac{n_2}{n_1} = \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \sqrt{2}. $

This indicates that the speed of light (and hence the wavelength) decreases by a factor of $\sqrt{2}$ as it enters the second medium. The frequency remains the same.

Therefore, the correct answer is

$ \lambda_2 = \frac{1}{\sqrt{2}} \lambda_1, \quad v_2 = v_1. $

$({\mu _1} - 1){A_1} = ({\mu _2} - 1){A_2}$

$ \Rightarrow (1.54 - 1)6 = (1.72 - 1){A_2}$

${A_2} = \left( {{{0.54} \over {0.72}} \times 6} \right) = {{18} \over 4} = \left( {{9 \over 2}} \right) = 4.5^\circ $

u = 25 cm

$f=\frac{1}{2}$ m = 50 cm

${1 \over v} - {1 \over u} = {1 \over f}$

$ \Rightarrow {1 \over v} + {1 \over {25}} = - {1 \over {50}}$

${1 \over v} = - {1 \over {50}}$

$ \Rightarrow u = - 50$ cm

Resolving power of microscope $ = \left( {{{2n\sin \theta } \over \lambda }} \right)$

$n\sin \theta $ = Numerical aperture

$n$ is the refractive index of medium.

The phenomenon described in the statement is called chromatic aberration.

Chromatic aberration is a type of optical aberration that occurs when a lens is unable to focus different colors of light at the same point after refraction. This causes the different colors of light to converge at different points on the principal axis, resulting in blurred or distorted images.

In the case of a convex lens, the refractive index of the lens is different for different colors of light. The blue light, having the shortest wavelength, refracts the most and the red light, having the longest wavelength, refracts the least. This causes the blue light to converge at a point closer to the lens than the red light, resulting in a blurred image.

Spherical aberration, on the other hand, is a type of optical aberration that occurs when a spherical lens is unable to focus all the incident light rays at a single point, resulting in a blurred image. Polarization refers to the orientation of the electric field of light waves, and scattering refers to the phenomenon of light being redirected in different directions due to interaction with matter.

Let, refractive index of medium = $\mu$

$\therefore$ $r + 15^\circ = 45^\circ $

$ \Rightarrow r = 30^\circ $

Using Snell's law,

$1\,.\,\sin 45^\circ = \sin 30^\circ \times \mu $

$ \Rightarrow {1 \over {\sqrt 2 }} = {1 \over 2} \times \mu $

$ \Rightarrow \mu = {2 \over {\sqrt 2 }} = \sqrt 2 = 1.414$

$\because$ ${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)$

$ \Rightarrow {1 \over {80}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right) \times {{(4 + 2)} \over {80}}$

$ \Rightarrow {{1.5} \over {{\mu _1}}} - 1 = {1 \over 6}$

$ \Rightarrow {{1.5} \over {{\mu _1}}} = {7 \over 6}$

$ \Rightarrow {\mu _1} = {{1.5 \times 6} \over 7} = {9 \over 7}$

$V = {1 \over {\sqrt {\mu \varepsilon } }} = {1 \over {\sqrt {{\mu _r}{\varepsilon _r}{\mu _0}{\varepsilon _0}} }}$

$ \Rightarrow {V_2} = {c \over {\sqrt 9 }} = {10^8}$ m/s

$\because$ $m = {{{f_o}} \over {{f_e}}}$

$ \Rightarrow 2 = {{{f_o}} \over {{f_e}}}$ ...... (i)

and, $l = {f_o} + {f_e}$

$ \Rightarrow 30 = {f_o} + {f_e}$ ..... (ii)

$ \Rightarrow 30 = {f_o} + {{{f_o}} \over 2}$

$ \Rightarrow 30 \times {2 \over 3} = {f_o}$

$ \Rightarrow {f_o} = 20$ cm

The resolving power of a microscope is determined by the Rayleigh criterion, which states that it is inversely proportional to the wavelength of light used in the medium in which the microscopy is being performed. Mathematically, the resolving power (RP) can be represented as:

$ RP = \frac{1}{\lambda_n} $

where $\lambda_n$ is the wavelength of light in the medium, which can be found using the formula:

$ \lambda_n = \frac{\lambda}{n} $

Here, $\lambda$ is the wavelength of light in vacuum (or air, for practical purposes, since their refractive indices are close enough), and $n$ is the refractive index of the medium.

In air, the refractive index $n = 1$, so the wavelength of light in air ($\lambda_{air}$) is equal to $\lambda$.

In oil, the refractive index $n = 2$, so the wavelength of light in oil ($\lambda_{oil}$) is $\lambda / 2$.

Therefore, the change in resolving power when moving from air to oil can be calculated as the ratio of resolving powers in oil to air:

$ \frac{RP_{oil}}{RP_{air}} = \frac{\lambda_{air}}{\lambda_{oil}} = \frac{\lambda}{\lambda / 2} = 2 $

This means that the resolving power in oil is twice that in air. Thus, the correct option is:

Option B: Resolving power will be twice in the oil than it was in the air.

Critical angle between them

$\sin {i_c} = {{{\mu _2}} \over {{\mu _1}}} = {{{v_1}} \over {{v_2}}}$

$\sin {i_c} = {3 \over 4}$

$ \Rightarrow \tan {i_c} = {3 \over {\sqrt 7 }}$

${i_c} = {\tan ^{ - 1}}{3 \over {\sqrt 7 }}$

The shift produced by the glass plate is

$d = t\left( {1 - {1 \over \mu }} \right) = 1 \times \left( {1 - {1 \over {1.5}}} \right) = {1 \over 3}$ cm

So final image must be produced at $\left( {12 - {1 \over 3}} \right)$ cm $ = {{35} \over 3}$ cm from lens so that glass plate must shift it to produce image at screen. So

${1 \over {12}} - {1 \over { - 240}} = {1 \over f} = {1 \over {35/3}} - {1 \over u}$

${1 \over u} = {3 \over {35}} - {1 \over {12}} - {1 \over {240}}$

or $u = - 560$ cm

so shift $ = 5.6 - 2.4 = 3.2$ m

In primary rainbow, observer sees red colour on the top and violet on the bottom.

${t_2} - {t_1} = 5 \times {10^{ - 10}}$

$ \Rightarrow {d \over {{v_B}}} - {d \over {{v_A}}} = 5 \times {10^{ - 10}}$

and, ${{{v_B}} \over {{v_A}}} = {{{\mu _A}} \over {{\mu _B}}} = {1 \over 2}$

$ \Rightarrow d\left( {1 - {{{v_B}} \over {{v_A}}}} \right) = 5 \times {10^{ - 10}} \times {v_B}$

$ \Rightarrow d\left( {1 - {1 \over 2}} \right) = 5 \times {10^{ - 10}} \times {v_B}$

$ \Rightarrow d = 10 \times {10^{ - 10}} \times {v_B}\,m$

$ \Rightarrow d = 5 \times {10^{ - 10}} \times {v_A}\,m$

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

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

For TIR

$\theta > {i_c}$

$\theta > {\sin ^{ - 1}}\left( {{{1.5} \over 2}} \right)$

$\theta > {\sin ^{ - 1}}\left( {{3 \over 4}} \right)$

$\mu = {{\sin \left( {{{{\delta _m} + A} \over 2}} \right)} \over {\sin (A/2)}} = \cot A/2$

$ \Rightarrow \cos A/2 = \sin \left( {{{{\delta _m} + A} \over 2}} \right)$

$ \Rightarrow {\pi \over 2} - {A \over 2} = {{{\delta _m} + A} \over 2}$

$ \Rightarrow \pi - 2A = {\delta _m}$

$R.P. = {1 \over {1.22\,\lambda /a}}$

$ = {{24.4 \times {{10}^{ - 2}}} \over {1.22 \times 2440 \times {{10}^{ - 10}}}}$

$ = 8.2 \times {10^5}$

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

${L_1}:{1 \over {{f_1}}} = (\mu - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) = {P_1} = (\mu - 1)\left( {{2 \over R}} \right) = P$

${L_2}:{1 \over {{f_2}}} = (\mu - 1)\left( {{1 \over {{R_1}}}} \right) = {P_2} = {{\mu - 1} \over R}$

${L_3}:{1 \over {{f_3}}} = (\mu - 1)\left( { - {1 \over {{R_2}}}} \right) = {P_3} = {{\mu - 1} \over R}$

According to the law,

$1 \times \sin \theta = \sqrt {2n} \times \sin \left( {{\theta \over 2}} \right)$

$ \Rightarrow \cos {\theta \over 2} = \sqrt {{n \over 2}} $

$ \Rightarrow \theta = 2{\cos ^{ - 1}}\left( {\sqrt {{n \over 2}} } \right)$

$n = \sqrt {K\mu } = 2$ (n $\Rightarrow$ refractive index)

So for TIR

$\theta > {\sin ^{ - 1}}\left( {{1 \over n}} \right)$

$\theta > 30$

Only option is 60$^\circ$

Speed of light in a medium $ = {c \over n}$

$\Rightarrow$ According to given information,

${c \over {{n_A}}} - {c \over {{n_B}}} = 2.6 \times {10^7}$

$ \Rightarrow {{{n_B}} \over {{n_A}}} - 1 = {{2.6 \times {{10}^7}} \over {3 \times {{10}^8}}} \times {n_B}$

$ \Rightarrow {{{n_B}} \over {{n_A}}} \simeq 1.13$