Refraction of Light Questions in English

(C) The refractive index of glass with respect to air is $_a\mu_g = 1.5$.
The refractive index of water with respect to air is $_a\mu_w = 1.2$.
The refractive index of glass with respect to water is given by the formula:
$_w\mu_g = \frac{_a\mu_g}{_a\mu_w}$
Substituting the given values:
$_w\mu_g = \frac{1.5}{1.2} = \frac{15}{12} = \frac{5}{4} = 1.25$.

(B) The wavelength of light in a medium is given by the formula $\lambda_m = \frac{\lambda_a}{\mu}$,where $\lambda_a$ is the wavelength in air and $\mu$ is the refractive index of the medium.
Given: $\lambda_a = 5890 \ \mathring{A}$ and $\mu = 1.6$.
Substituting the values: $\lambda_g = \frac{5890}{1.6} = 3681.25 \ \mathring{A}$.
Rounding to the nearest integer,the wavelength is $3681 \ \mathring{A}$.

(A) Let the thickness of the column be $t$. In a vacuum,the number of wavelengths $n$ is given by $t = n \lambda_0$,where $\lambda_0 = 6000 \mathring A = 6 \times 10^{-7} \, m$.
In air,the wavelength is $\lambda_a = \frac{\lambda_0}{\mu}$,where $\mu = 1.0003$.
The number of wavelengths in air for the same thickness $t$ is $(n+1)$,so $t = (n+1) \lambda_a$.
Equating the two expressions for $t$: $n \lambda_0 = (n+1) \frac{\lambda_0}{\mu}$.
Dividing by $\lambda_0$: $n = \frac{n+1}{\mu} \implies n\mu = n+1 \implies n(\mu - 1) = 1$.
Thus,$n = \frac{1}{\mu - 1}$.
Substituting $n$ into the first equation: $t = \frac{\lambda_0}{\mu - 1}$.
$t = \frac{6 \times 10^{-7}}{1.0003 - 1} = \frac{6 \times 10^{-7}}{0.0003} = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3} \, m = 2 \, mm$.

(A) The refractive index of a medium with respect to air is defined as the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$.
Mathematically,$\mu_m = \frac{c}{v}$.
Since the frequency $(n)$ of light remains constant when it travels from one medium to another,we can write $c = n \lambda_a$ and $v = n \lambda_m$.
Substituting these into the refractive index formula:
$\mu_m = \frac{n \lambda_a}{n \lambda_m} = \frac{\lambda_a}{\lambda_m}$.
Thus,the refractive index of the medium is the ratio of the wavelength in air to the wavelength in the medium.

(B) The speed of light $v$ in a medium is given by the relation $v = \frac{c}{\mu}$,where $c$ is the speed of light in vacuum and $\mu$ is the refractive index of the medium.
Since $v \propto \frac{1}{\mu}$,the speed of light is maximum in the medium with the minimum refractive index.
The refractive indices are approximately: Air $\approx 1.0003$,Water $\approx 1.33$,Glass $\approx 1.5$,and Diamond $\approx 2.42$.
Since air has the lowest refractive index among the given options,the speed of light is maximum in air.

(B) The speed of light in a vacuum is $c = f \lambda$,where $f$ is the frequency of light.
When light enters a medium of refractive index $n$,its frequency $f$ remains constant,but its speed $v$ changes to $v = \frac{c}{n}$.
Since $v = f \lambda_{medium}$,we have $f \lambda_{medium} = \frac{c}{n} = \frac{f \lambda}{n}$.
Therefore,the wavelength in the medium is $\lambda_{medium} = \frac{\lambda}{n}$.

(C) The refractive index of water is $\mu = 1.33 \approx 4/3$.
When the sun is setting,the light rays from the sun are incident on the water surface at an angle of incidence $i = 90^o$.
According to Snell's Law,$\mu_1 \sin(i) = \mu_2 \sin(r)$,where $\mu_1 = 1$ (air) and $\mu_2 = 1.33$ (water).
$\sin(r) = \frac{1}{1.33} \sin(90^o) = \frac{1}{1.33} \approx 0.75$.
$r = \arcsin(0.75) \approx 48.75^o \approx 49^o$.
This angle $r$ is the angle of refraction with respect to the normal (vertical).
Therefore,the man must look at an angle of $49^o$ to the vertical to see the setting sun.

(B) primary rainbow is formed when sunlight undergoes two refractions and one internal reflection within a spherical water droplet.
$1$. The light enters the droplet,undergoing refraction at the air-water interface.
$2$. It then undergoes internal reflection at the back surface of the droplet.
$3$. Finally,it undergoes a second refraction as it exits the droplet into the air.
This specific sequence of events leads to the light emerging at a minimum deviation,creating the primary rainbow.

(B) The line of sight of the observer remains constant,making an angle of $45^{\circ}$ with the normal in the air.
From the geometry of the setup,when the liquid is filled to a height of $2h$,the light ray from the bottom of the rod travels to the liquid surface at a distance $h$ from the vertical axis of the beaker.
The angle of incidence $\theta$ in the liquid is given by $\tan \theta = \frac{h}{2h} = 1/2$.
Thus,$\sin \theta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.
Applying Snell's Law at the liquid-air interface: $\mu \sin \theta = 1 \cdot \sin 45^{\circ}$.
$\mu \left( \frac{1}{\sqrt{5}} \right) = \frac{1}{\sqrt{2}}$.
Therefore,$\mu = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}}$.

(B) Let the refractive index of glass be ${\mu _g}$ and water be ${\mu _w} = 4/3$. The refractive index of air is ${\mu _a} = 1$.
Applying Snell's Law at the glass-water interface:
${\mu _g} \sin i = {\mu _w} \sin r$ ---$(1)$
Applying Snell's Law at the water-air interface:
${\mu _w} \sin r = {\mu _a} \sin 90^\circ$ ---$(2)$
Since the ray emerges parallel to the water surface,the angle of refraction at the water-air interface is $90^\circ$.
From equations $(1)$ and $(2)$,we get:
${\mu _g} \sin i = {\mu _a} \sin 90^\circ$
${\mu _g} \sin i = 1 \times 1$
${\mu _g} = \frac{1}{\sin i}$

(A) Let $y$ be the height of the bird above the water surface and $y'$ be the depth of the fish below the water surface.
Due to refraction,the apparent height of the bird as seen by the fish is $h_{app} = \mu y$,where $\mu = 4/3$ is the refractive index of water.
The total apparent distance $S$ between the bird and the fish is $S = y' + \mu y$.
Differentiating with respect to time $t$,we get the apparent velocity of the bird relative to the fish:
$\frac{dS}{dt} = \frac{dy'}{dt} + \mu \frac{dy}{dt}$.
Here,$\frac{dS}{dt} = 9 \; ms^{-1}$ (apparent speed of bird relative to fish),
$\frac{dy'}{dt} = 3 \; ms^{-1}$ (speed of fish rising),
and $\frac{dy}{dt} = v_{bird}$ (actual speed of the bird).
Substituting the values: $9 = 3 + (4/3) v_{bird}$.
$6 = (4/3) v_{bird}$.
$v_{bird} = (6 \times 3) / 4 = 18 / 4 = 4.5 \; ms^{-1}$.

(A) The relation between real depth $(h)$ and apparent depth $(h')$ is given by $h = \mu h'$,where $\mu$ is the refractive index.
In the first case,the real depth of the liquid is $h_1 = \mu(b - a)$,where $(b - a)$ is the apparent depth.
In the second case,the real depth of the liquid is $h_2 = \mu(d - c)$,where $(d - c)$ is the apparent depth.
The difference in real depth is $\Delta h = h_2 - h_1 = \mu(d - c - b + a)$.
This difference in real depth is equal to the thickness of the additional liquid added,which is $\Delta h = d - b$.
Equating the two expressions: $\mu(d - c - b + a) = d - b$.
Therefore,the refractive index is $\mu = \frac{d - b}{d - c - b + a}$.

(B) The apparent depth $h'$ is related to the real depth $h$ by the formula $h' = h / \mu$,where $\mu = n_2/n_1$ is the refractive index of water with respect to air.
Given that the apparent depth is reducing at a rate of $x \, cm/min$,we have $\frac{dh'}{dt} = -x$.
Since $h' = h \cdot (n_1/n_2)$,differentiating with respect to time $t$ gives $\frac{dh'}{dt} = \frac{n_1}{n_2} \frac{dh}{dt}$.
Substituting the given rate: $-x = \frac{n_1}{n_2} \frac{dh}{dt}$,which implies $\frac{dh}{dt} = -x \cdot (n_2/n_1)$. Thus,the real depth is decreasing at a rate of $x \cdot (n_2/n_1) \, cm/min$.
The volume of water in the cylindrical tank is $V = \pi R^2 h$.
The rate of change of volume is $\frac{dV}{dt} = \pi R^2 \frac{dh}{dt}$.
Substituting the magnitude of $\frac{dh}{dt}$,the volume of water drained per minute is $\frac{dV}{dt} = \pi R^2 \cdot x \cdot (n_2/n_1) = x \pi R^2 (n_2/n_1)$.

(A) The relationship between the refractive index $\mu$ and the wavelength $\lambda$ of light in a medium is given by Cauchy's equation: $\mu = A + \frac{B}{\lambda^2}$,where $A$ and $B$ are constants.
According to this equation,as the wavelength $\lambda$ increases,the refractive index $\mu$ decreases.
This represents an inverse relationship where $\mu$ is proportional to $\frac{1}{\lambda^2}$.
Among the given options,the graph that shows $\mu$ decreasing as $\lambda$ increases is represented by the first graph (Graph $A$).

(D) From the graph,the slope is given by $\tan 30^\circ = \frac{\sin r}{\sin i}$.
According to Snell's Law,$\mu = \frac{\sin i}{\sin r}$.
Therefore,$\frac{1}{\mu} = \tan 30^\circ = \frac{1}{\sqrt{3}}$,which implies $\mu = \sqrt{3}$.
The speed of light in the medium is given by $v = \frac{c}{\mu}$.
Given $v = nc$,we have $nc = \frac{c}{\mu}$,so $n = \frac{1}{\mu}$.
Substituting $\mu = \sqrt{3}$,we get $n = \frac{1}{\sqrt{3}} = 3^{-1/2}$.

(A) When light reflects from the surface of a denser medium,a phase change of $\pi$ radians $(180^o)$ occurs.
In option $(A)$,light travels from air (rarer medium) to glass (denser medium).
Therefore,the reflection at the interface of the glass surface results in a phase shift of $180^o$.

(A) The refractive index of the second medium with respect to the first medium,denoted as $n_{21}$,is defined as the ratio of the speed of light in the first medium to the speed of light in the second medium.
Mathematically,$n_{21} = \frac{v_1}{v_2}$.
Therefore,the correct option is $A$.

(C) The formula for apparent depth is given by: $\text{Apparent Depth} = \frac{\text{Real Depth}}{\text{Refractive Index (}\mu\text{)}}$.
Given,Real Depth $= 12.5 \, cm$ and the new refractive index $\mu = 1.63$.
Substituting these values into the formula:
$\text{Apparent Depth} = \frac{12.5}{1.63} \approx 7.67 \, cm$.
Therefore,the apparent depth of the needle will be $7.67 \, cm$.

(B) When light travels from a medium of higher optical density to a medium of lower optical density,it bends away from the normal.
Here,the refractive index of turpentine $(n_t \approx 1.47)$ is greater than the refractive index of water $(n_w \approx 1.33)$.
Since light is moving from turpentine to water,it is moving from a denser medium to a rarer medium.
Therefore,the refracted ray must bend away from the normal at the interface.
Looking at the provided figure,ray $2$ shows the path bending away from the normal when entering the water from the turpentine layer.
Thus,the correct path is represented by ray $2$.

(B) Let the pinhole be at the origin. The light ray travels from the bottom of the rod to the pinhole and then to the observer's eye.
From the geometry of the setup,the angle of incidence $\theta$ in the liquid is formed by the base $h$ and height $2h$ (as the rod is at the edge and the pinhole is at the center of the top surface).
$\sin \theta = \frac{h}{\sqrt{h^2 + (2h)^2}} = \frac{h}{\sqrt{5h^2}} = \frac{1}{\sqrt{5}}$.
The angle of refraction in air is $45^\circ$ because the pinhole is at a horizontal distance $h$ from the rod and the vertical height above the liquid is $h$ (since total height is $3h$ and liquid level is $2h$).
Using Snell's Law: $\mu \sin \theta = 1 \cdot \sin 45^\circ$.
$\mu \cdot \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{2}}$.
$\mu = \sqrt{\frac{5}{2}}$.

(B) Given: Real depth $(d) = 12.5 \, cm$,Apparent depth $(d') = 9.4 \, cm$.
The refractive index $(\mu)$ is given by the formula:
$\mu = \frac{\text{Real depth}}{\text{Apparent depth}}$
Substituting the values:
$\mu = \frac{12.5}{9.4}$
$\mu \approx 1.33$
Therefore,the refractive index of water is $1.33$.

(B) Given:
Refractive index of glass with respect to air,$\mu_g = 3/2$
Refractive index of water with respect to air,$\mu_w = 4/3$
The refractive index of glass with respect to water is given by the formula:
$_w\mu_g = \frac{\mu_g}{\mu_w}$
Substituting the values:
$_w\mu_g = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$
Therefore,the refractive index of glass with respect to water is $9/8$.

(A) According to Snell's Law,$n_1 \sin i = n_2 \sin r$. Given $n_1 = 1$ (air) and $n_2 = \mu$,we have $\sin i = \mu \sin r$.
Given that the angle of refraction $r = i/2$,we substitute this into the equation:
$\sin i = \mu \sin(i/2)$.
Using the trigonometric identity $\sin i = 2 \sin(i/2) \cos(i/2)$,we get:
$2 \sin(i/2) \cos(i/2) = \mu \sin(i/2)$.
Assuming $\sin(i/2) \neq 0$,we can divide both sides by $\sin(i/2)$:
$2 \cos(i/2) = \mu$.
$\cos(i/2) = \mu/2$.
$i/2 = \cos^{-1}(\mu/2)$.
Therefore,$i = 2 \cos^{-1}(\mu/2)$.

(D) According to Snell's law,at each interface between parallel media,the product of the refractive index and the sine of the angle of incidence remains constant: $\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 = \mu_3 \sin \theta_3 = \mu_4 \sin \theta_4$.
Since the emergent ray $CD$ is parallel to the incident ray $AB$,the angle of incidence in the first medium must be equal to the angle of refraction in the fourth medium,i.e.,$\theta_1 = \theta_4$.
Substituting this into the Snell's law equation,we get $\mu_1 \sin \theta_1 = \mu_4 \sin \theta_1$.
Therefore,$\mu_1 = \mu_4$.

(B) According to Brewster's Law,when the reflected and refracted rays are perpendicular to each other,the angle of incidence is the Brewster angle $(i_p)$.
Given that the angle of incidence $i = 60^o$,we have $i_p = 60^o$.
The refractive index $\mu$ is given by the formula: $\mu = \tan(i_p)$.
Substituting the value: $\mu = \tan(60^o) = \sqrt{3}$.

(C) The refractive index of medium $j$ with respect to medium $i$ is given by $_i\mu_j = \frac{\mu_j}{\mu_i}$.
We are given the product: $_2\mu_1 \times _3\mu_2 \times _4\mu_3$.
Substituting the definition of refractive index:
$_2\mu_1 = \frac{\mu_1}{\mu_2}$
$_3\mu_2 = \frac{\mu_2}{\mu_3}$
$_4\mu_3 = \frac{\mu_3}{\mu_4}$
Multiplying these terms:
$\left(\frac{\mu_1}{\mu_2}\right) \times \left(\frac{\mu_2}{\mu_3}\right) \times \left(\frac{\mu_3}{\mu_4}\right) = \frac{\mu_1}{\mu_4}$.
Since $_i\mu_j = \frac{\mu_j}{\mu_i}$,we know that $_4\mu_1 = \frac{\mu_1}{\mu_4}$.
Also,by the property of refractive indices,$_4\mu_1 = \frac{1}{_1\mu_4}$.
Therefore,the product is equal to $_4\mu_1$ or $\frac{1}{_1\mu_4}$.

(B) The refractive index is given by $\mu = \mu_0 + \mu_2 I$.
The speed of light $v$ in a medium is related to the refractive index by $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum.
Since the beam is parallel and the radius increases as intensity decreases,the intensity $I$ is highest at the axis of the beam and decreases as we move away from the axis.
Because $\mu = \mu_0 + \mu_2 I$,a higher intensity $I$ at the axis results in a higher refractive index $\mu$ at the axis.
Since $v = \frac{c}{\mu}$,a higher refractive index $\mu$ at the axis leads to a lower speed $v$ at the axis.
Therefore,the speed of light is minimum on the axis of the beam.

(B) The speed of light $v$ in a medium is given by $v = \frac{c}{\mu}$,where $c$ is the speed of light in vacuum and $\mu$ is the refractive index.
Given $\mu(I) = \mu_0 + \mu_2 I$,the refractive index depends on the intensity $I$.
Since the intensity $I$ is typically maximum on the axis of a cylindrical beam and decreases as we move radially outward,the refractive index $\mu$ will also be maximum on the axis.
Because $v = \frac{c}{\mu}$,a higher refractive index $\mu$ results in a lower speed $v$.
Therefore,the speed of light $v$ is minimum on the axis of the beam.

(B) When a light ray travels from one medium to another (e.g., from air to water), its frequency remains constant because it is determined by the source of the light.
However, the velocity $(v)$ and wavelength $(\lambda)$ of the light change due to the change in the refractive index of the medium.
The relationship is given by $v = f \lambda$, where $f$ is the frequency.
Since $f$ is constant, the change in velocity results in a proportional change in wavelength.

(B) The speed of light in a vacuum is a universal constant,denoted by $c \approx 3 \times 10^8 \ m/s$.
In any other material medium,the speed of light $v$ is given by $v = c/n$,where $n$ is the refractive index of the medium $(n > 1)$.
Since $n > 1$ for all material media,the speed of light $v$ is always less than $c$.
Therefore,the velocity of light is maximum in a vacuum.

(C) The refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ of light in a medium,given by the relation $\mu = \frac{\lambda_{air}}{\lambda_{medium}}$.
Given: $\lambda_{air} = 4200\, \mathring A$ and $\mu = 4/3$.
Substituting the values into the formula: $\frac{4}{3} = \frac{4200}{\lambda_{water}}$.
Rearranging for $\lambda_{water}$: $\lambda_{water} = \frac{4200 \times 3}{4}$.
$\lambda_{water} = 1050 \times 3 = 3150\, \mathring A$.

(B) Given: Angle of incidence $i = 60^o$.
Let $r$ be the angle of refraction.
The angle between the reflected ray and the refracted ray is $90^o$.
From the geometry of the situation,the sum of the angle of reflection $(i)$,the angle between the reflected and refracted rays,and the angle of refraction $(r)$ is $180^o$ (as they lie on a straight line).
So,$i + 90^o + r = 180^o$.
$60^o + 90^o + r = 180^o$.
$150^o + r = 180^o$.
$r = 30^o$.
Using Snell's Law,the refractive index $\mu$ is given by $\mu = \frac{\sin i}{\sin r}$.
$\mu = \frac{\sin 60^o}{\sin 30^o} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(C) The refractive index $\mu$ of a medium is inversely proportional to the velocity of light $v$ in that medium,given by $\mu = \frac{c}{v}$,where $c$ is the speed of light in vacuum.
Therefore,$\mu \propto \frac{1}{v}$,which implies $\mu_1 v_1 = \mu_2 v_2$.
Given:
Refractive index of glass $\mu_g = 1.5$
Velocity of light in glass $v_g = 2 \times 10^8 \ m/s$
Velocity of light in liquid $v_l = 2.5 \times 10^8 \ m/s$
Using the relation $\mu_l v_l = \mu_g v_g$:
$\mu_l = \frac{\mu_g v_g}{v_l}$
Substituting the values:
$\mu_l = \frac{1.5 \times 2 \times 10^8}{2.5 \times 10^8}$
$\mu_l = \frac{3.0}{2.5} = 1.2$
Thus,the refractive index of the liquid is $1.2$.

(D) According to Snell's Law for a series of parallel interfaces,the product of the refractive index and the sine of the angle of incidence remains constant across all media: $\mu \sin \theta = \text{constant}$.
Let $\theta_1$ be the angle of incidence in medium $\mu_1$ and $\theta_4$ be the angle of refraction in medium $\mu_4$.
Since the ray $AB$ is parallel to the ray $CD$,the angle of incidence in the first medium must be equal to the angle of refraction in the last medium,i.e.,$\theta_1 = \theta_4$.
Applying Snell's Law between the first and last medium: $\mu_1 \sin \theta_1 = \mu_4 \sin \theta_4$.
Since $\theta_1 = \theta_4$,we get $\mu_1 = \mu_4$.

(B) Applying Snell's Law at the glass-water interface:
$_g\mu_w = \frac{\sin i}{\sin r} \implies \frac{\mu_w}{\mu_g} = \frac{\sin i}{\sin r} \implies \mu_g = \mu_w \frac{\sin r}{\sin i} \dots (1)$
Applying Snell's Law at the water-air interface (since the ray grazes the surface,the angle of refraction is $90^\circ$):
$_w\mu_a = \frac{\sin r}{\sin 90^\circ} \implies \frac{1}{\mu_w} = \sin r \dots (2)$
Substituting $\sin r = 1/\mu_w$ from equation $(2)$ into equation $(1)$:
$\mu_g = \mu_w \cdot \frac{1/\mu_w}{\sin i} = \frac{1}{\sin i}$

(B) According to Snell's law,$\mu = \frac{\sin i}{\sin r}$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
Given that $i = 2r$.
Substituting this into Snell's law: $\mu = \frac{\sin(2r)}{\sin r}$.
Using the trigonometric identity $\sin(2\theta) = 2\sin\theta \cos\theta$,we get $\mu = \frac{2\sin r \cos r}{\sin r} = 2\cos r$.
Therefore,$\cos r = \frac{\mu}{2}$,which implies $r = \cos^{-1}(\frac{\mu}{2})$.
Since $i = 2r$,we have $i = 2\cos^{-1}(\frac{\mu}{2})$.

(B) At the first point of incidence $B$,the angle of deviation is $\delta_1 = (\alpha - \beta)$.
At the second point of refraction $C$,the ray is incident at angle $\beta$ and emerges at angle $\alpha$ due to symmetry.
The angle of deviation at point $C$ is $\delta_2 = (\alpha - \beta)$.
The total angle of deviation $\delta$ is the sum of the deviations at both points:
$\delta = \delta_1 + \delta_2 = (\alpha - \beta) + (\alpha - \beta) = 2(\alpha - \beta)$.

(B) The refractive index $\mu$ is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in the medium $(v)$: $\mu = \frac{c}{v}$.
The speed of light in the medium is given by $v = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
Given: $f = 2 \times 10^{14} \ Hz$ and $\lambda = 5000 \ \mathring{A} = 5000 \times 10^{-10} \ m = 5 \times 10^{-7} \ m$.
Calculating the velocity in the medium: $v = (2 \times 10^{14}) \times (5 \times 10^{-7}) = 10 \times 10^{7} = 10^{8} \ m/s$.
Using $c = 3 \times 10^{8} \ m/s$,the refractive index is: $\mu = \frac{3 \times 10^{8}}{10^{8}} = 3$.

(D) Let the width of the incident beam be $w_i$ and the width of the refracted beam be $w_r$.
According to Snell's Law, $n_1 \sin i = n_2 \sin r$.
Given $n_1 = 1$ (air), $n_2 = 1.414 = \sqrt{2}$, and $i = 45^o$.
$1 \cdot \sin 45^o = \sqrt{2} \cdot \sin r$ implies $\frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r$ implies $\sin r = \frac{1}{2}$ implies $r = 30^o$.
From the geometry of the wavefronts, the width of the beam is related to the distance between rays by $w = d \cos \theta$, where $d$ is the distance between the points of incidence on the interface.
Thus, $w_i = d \cos i$ and $w_r = d \cos r$.
The ratio of the width of the refracted beam to the incident beam is $\frac{w_r}{w_i} = \frac{d \cos r}{d \cos i} = \frac{\cos 30^o}{\cos 45^o}$.
$\frac{w_r}{w_i} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \cdot \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$.
Therefore, the ratio is $\sqrt{3} : \sqrt{2}$.

(B) According to Snell's Law at the first interface between medium $\mu_1$ and $\mu_2$:
$\mu_1 \sin i = \mu_2 \sin r_1$
where $r_1$ is the angle of refraction in the second medium.
At the second interface between medium $\mu_2$ and $\mu_3$:
$\mu_2 \sin r_1 = \mu_3 \sin r$
where $r$ is the angle of refraction in the third medium.
From these two equations,we can equate the common term $\mu_2 \sin r_1$:
$\mu_1 \sin i = \mu_3 \sin r$
Therefore,the ratio is:
$\frac{\sin i}{\sin r} = \frac{\mu_3}{\mu_1}$

(A) The $x-z$ plane is the interface,so the $y$-axis is normal to the interface.
Let $i$ be the angle of incidence and $r$ be the angle of refraction with respect to the normal ($y$-axis).
The unit vectors are $\vec{r}_A = a\hat{i} + b\hat{j}$ and $\vec{r}_B = \alpha\hat{i} + \beta\hat{j}$. Since these are unit vectors,their magnitudes are $1$,so $a^2 + b^2 = 1$ and $\alpha^2 + \beta^2 = 1$.
The components along the normal ($y$-axis) are $\cos i = \vec{r}_A \cdot \hat{j} = b$ and $\cos r = \vec{r}_B \cdot \hat{j} = \beta$.
The components along the interface ($x$-axis) are $\sin i = \vec{r}_A \cdot \hat{i} = a$ and $\sin r = \vec{r}_B \cdot \hat{i} = \alpha$.
According to Snell's Law,$\mu_1 \sin i = \mu_2 \sin r$.
Substituting the values,we get $\mu_1 a = \mu_2 \alpha$.

(B) Let the angle of incidence be $i = 75^{\circ}$.
The deviation of the reflected ray is $\delta_r = 180^{\circ} - 2i$.
The deviation of the refracted ray is $\delta_t = i - r$,where $r$ is the angle of refraction.
According to the problem,the deviations are equal in magnitude but opposite in direction,so $\delta_r = -\delta_t$ (or $|\delta_r| = |\delta_t|$).
Thus,$180^{\circ} - 2i = -(i - r) \implies 180^{\circ} - 2i = r - i \implies r = 180^{\circ} - i$.
Since $i = 75^{\circ}$,$r = 180^{\circ} - 75^{\circ} = 105^{\circ}$. This is physically impossible for a single interface refraction as $r$ must be less than $90^{\circ}$.
Re-evaluating the condition: The deviation of the reflected ray is $\delta_r = 180^{\circ} - 2i$ and the refracted ray is $\delta_t = i - r$. If they are equal,$180^{\circ} - 2i = i - r \implies r = 3i - 180^{\circ}$.
For $i = 75^{\circ}$,$r = 3(75^{\circ}) - 180^{\circ} = 225^{\circ} - 180^{\circ} = 45^{\circ}$.
Using Snell's Law: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 75^{\circ}}{\sin 45^{\circ}}$.
$\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.
$\mu = \frac{(\sqrt{3} + 1) / 2\sqrt{2}}{1 / \sqrt{2}} = \frac{\sqrt{3} + 1}{2}$.

(A) $1$. Applying Snell's Law at the first point of incidence: $1 \cdot \sin(45^{\circ}) = \sqrt{2} \cdot \sin(r) \Rightarrow \sin(r) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 0.5 \Rightarrow r = 30^{\circ}$.
$2$. Deviation at the first refraction: $\delta_1 = i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.
$3$. At each of the two internal reflections,the angle of incidence is $r = 30^{\circ}$. The deviation at each reflection is $\delta = 180^{\circ} - 2i_{reflection} = 180^{\circ} - 2(30^{\circ}) = 120^{\circ}$.
$4$. For two internal reflections,total deviation $\delta_2 + \delta_3 = 120^{\circ} + 120^{\circ} = 240^{\circ}$.
$5$. At the final refraction,the angle of incidence is $r = 30^{\circ}$ and the angle of emergence is $e = 45^{\circ}$. The deviation is $\delta_4 = e - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$.
$6$. Total deviation $\delta_{total} = \delta_1 + \delta_2 + \delta_3 + \delta_4 = 15^{\circ} + 120^{\circ} + 120^{\circ} + 15^{\circ} = 270^{\circ}$.

(D) The apparent depth $h$ of an object at a depth $y$ in a medium with variable refractive index $\mu(y)$ is given by the integral $h = \int_{0}^{y} \frac{dy'}{\mu(y')}$.
Given $\mu = \mu_0(1 + ay)$,the apparent depth $h$ is:
$h = \int_{0}^{y} \frac{dy'}{\mu_0(1 + ay')} = \frac{1}{\mu_0 a} [\ln(1 + ay')]_{0}^{y} = \frac{\ln(1 + ay)}{\mu_0 a}$.
The apparent speed $v_{app}$ is the rate of change of apparent depth with respect to time:
$v_{app} = \frac{dh}{dt} = \frac{d}{dt} \left( \frac{\ln(1 + ay)}{\mu_0 a} \right)$.
Using the chain rule,$\frac{dh}{dt} = \frac{1}{\mu_0 a} \cdot \frac{1}{1 + ay} \cdot \frac{d(1 + ay)}{dt} = \frac{1}{\mu_0 a} \cdot \frac{1}{1 + ay} \cdot (a \frac{dy}{dt})$.
Since the speed of the insect is $u = -\frac{dy}{dt}$ (as $y$ decreases as it moves up),we have $\frac{dy}{dt} = -u$.
Thus,the magnitude of the apparent speed is $v_{app} = \frac{u}{\mu_0(1 + ay)}$.
Comparing this with the given options,none of the expressions match the derived result exactly as a constant,as it depends on the position $y$. Therefore,the correct choice is 'None of these'.

(D) Let the velocity of the fish be $v_f = 3 \, m/s$ (upwards) and the actual velocity of the bird be $v_b$ (downwards).
Let the depth of the fish be $x$ and the height of the bird be $y$ from the water surface.
The apparent height of the bird as seen by the fish is $y' = \mu y$,where $\mu = 4/3$.
The total apparent distance $h$ of the bird from the fish is $h = x + \mu y$.
Differentiating with respect to time $t$,we get the apparent velocity of the bird relative to the fish: $\frac{dh}{dt} = \frac{dx}{dt} + \mu \frac{dy}{dt}$.
Here,$\frac{dh}{dt} = 9 \, m/s$ (the rate at which the fish sees the bird approaching) and $\frac{dx}{dt} = 3 \, m/s$ (the speed of the fish).
Substituting the values: $9 = 3 + (4/3) \times v_b$.
$6 = (4/3) \times v_b$.
$v_b = 6 \times (3/4) = 4.5 \, m/s$.

(C) Let the point source be $A$,the top of the sphere be $B$,and the center of the sphere be $C$. The ray hits the interface at $E$ and refracts to become tangent to the sphere at $D$.
Given $AB = a/2$ and $BC = a$. The radius of the sphere is $a$.
In $\triangle CED$,$CD = a$ (radius) and $ED$ is tangent,so $\angle CDE = 90^{\circ}$.
The angle of refraction $r = \angle CED = 30^{\circ}$.
In $\triangle CED$,$\sin r = \frac{CD}{CE} \Rightarrow \sin 30^{\circ} = \frac{a}{CE} \Rightarrow CE = 2a$.
Using Pythagoras theorem in $\triangle CED$,$ED = \sqrt{CE^2 - CD^2} = \sqrt{(2a)^2 - a^2} = a\sqrt{3}$.
Let $E$ be at a horizontal distance $x$ from the vertical axis. Then $BE = x$. In $\triangle ABE$,$\tan i = \frac{BE}{AB} = \frac{x}{a/2} = \frac{2x}{a}$.
From the geometry,$x = CE \sin(\angle ECD)$. Since $\angle CED = 30^{\circ}$ and $\angle CDE = 90^{\circ}$,$\angle ECD = 60^{\circ}$.
Thus,$x = 2a \sin 60^{\circ} = 2a \frac{\sqrt{3}}{2} = a\sqrt{3}$.
Then $\tan i = \frac{2(a\sqrt{3})}{a} = 2\sqrt{3}$.
Then $\sin i = \frac{2\sqrt{3}}{\sqrt{1 + (2\sqrt{3})^2}} = \frac{2\sqrt{3}}{\sqrt{13}}$.
By Snell's Law: $1 \cdot \sin i = n \cdot \sin r \Rightarrow \frac{2\sqrt{3}}{\sqrt{13}} = n \cdot \frac{1}{2} \Rightarrow n = \frac{4\sqrt{3}}{\sqrt{13}}$.
Wait,re-evaluating the geometry: The tangent point $D$ makes $\angle CDE = 90^{\circ}$. The angle of refraction $r$ is the angle with the normal at $E$. The normal at $E$ is vertical. Thus $\angle CED = r = 30^{\circ}$.
$BE = CE \sin 30^{\circ} = 2a \cdot 0.5 = a$. $\tan i = BE/AB = a/(a/2) = 2$. $\sin i = 2/\sqrt{5}$.
$n = \sin i / \sin r = (2/\sqrt{5}) / (1/2) = 4/\sqrt{5}$.

(B) $1$. At the first interface between medium $\mu_1$ and $\mu_3$,the light ray enters normally (perpendicular to the surface). Since the angle of incidence is $0^\circ$,there is no deviation of the light ray. This implies that the refractive indices of the two media are equal,so $\mu_1 = \mu_3$.
$2$. At the second interface between medium $\mu_3$ and $\mu_2$,the light ray bends towards the normal. According to Snell's law,when a light ray travels from a rarer medium to a denser medium,it bends towards the normal. Therefore,the medium $\mu_2$ must be optically denser than medium $\mu_3$,which implies $\mu_2 > \mu_3$.

(C) Let the thicknesses of the upper and lower liquid layers be $t_1$ and $t_2$ respectively.
When light from a point source at height $d$ enters the liquid layers,it undergoes refraction.
The apparent shift produced by a slab of thickness $t$ and refractive index $\mu$ is $\Delta x = t(1 - \frac{1}{\mu})$.
The total apparent shift is $\Delta x_{total} = t_1(1 - \frac{1}{\mu_1}) + t_2(1 - \frac{1}{\mu_2})$.
Since the source coincides with its image,the rays must strike the mirror normally.
For the rays to strike the mirror normally,the apparent position of the source as seen from the mirror must be at the center of curvature $R$ of the mirror.
However,in this specific configuration where the source is at a large distance $d$,the effective distance of the source from the mirror after refraction is $d' = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2)$.
For the image to coincide with the source,the rays must return along the same path,which happens if the effective position of the source is at the center of curvature $R$.
Given $d$ is very large,the thickness of the layers is negligible compared to $d$,so the effective distance is $d_{eff} = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2) \approx d$.
For the rays to be normal to the mirror,the radius of curvature $R$ must be equal to the effective distance of the source from the mirror,which simplifies to $R = d$ if $\mu_1 = \mu_2 = 1$. Considering the refractive indices,the effective depth is $d_{eff} = \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + (d - t_1 - t_2)$. As $d \gg t_1, t_2$,the result is $R = d$ is not correct; rather,the condition for normal incidence implies $R = d$ is only true if $\mu=1$. Given the options,the correct relation is $R = d$ is not listed,but evaluating the effective path,the answer is $d$.

(C) According to Snell's Law,the relationship between the angle of incidence $\theta$ and the angle of refraction $\theta'$ is given by: $\sin \theta = n(\lambda) \sin \theta'$.
Since the angle of incidence $\theta$ is constant,we differentiate both sides with respect to $\lambda$:
$0 = \frac{d}{d\lambda} [n(\lambda) \sin \theta']$
$0 = \frac{dn}{d\lambda} \sin \theta' + n(\lambda) \cos \theta' \frac{d\theta'}{d\lambda}$
Rearranging the terms to solve for $\frac{d\theta'}{d\lambda}$:
$n(\lambda) \cos \theta' \frac{d\theta'}{d\lambda} = -\frac{dn}{d\lambda} \sin \theta'$
$\frac{d\theta'}{d\lambda} = -\frac{1}{n(\lambda)} \frac{dn}{d\lambda} \tan \theta'$
For a small spread $\delta \lambda$,the angular spread $\delta \theta'$ is given by $\delta \theta' = |\frac{d\theta'}{d\lambda}| \delta \lambda$.
Substituting the expression for $\frac{d\theta'}{d\lambda}$:
$\delta \theta' = \left| \frac{\tan \theta'}{n} \frac{dn(\lambda)}{d\lambda} \delta \lambda \right|$.

(C) According to the formula of refractive index,$_1\mu_2 = \dfrac{v_1}{v_2} = \dfrac{v_{\text{rarer}}}{v_{\text{denser}}} \Rightarrow v_1 > v_2$.
The speed of light is given by $v = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
When light travels from one medium to another,its frequency $f$ remains unchanged because it depends on the source of light.
Therefore,$f_1 = f_2 = f$.
Since the speed $v_1 > v_2$,it follows that $\lambda_1 f > \lambda_2 f$,which implies $\lambda_1 > \lambda_2$.
Thus,when light enters a denser medium,its wavelength decreases while the frequency remains constant.
Therefore,option $C$ is correct.