Refraction of Light Questions in English

(D) According to Snell's Law for waves,the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities in the respective media:
$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$
Given:
Angle of incidence $i = 30^\circ$
Velocity in medium $P$ $(v_1)$ = $u$
Velocity in medium $Q$ $(v_2)$ = $2u$
Substituting the values into the formula:
$\frac{\sin 30^\circ}{\sin r} = \frac{u}{2u}$
$\frac{0.5}{\sin r} = \frac{1}{2}$
$\sin r = 0.5 \times 2 = 1$
Since $\sin r = 1,$ the angle of refraction $r = 90^\circ.$

(A) The refractive index of a medium depends on the wavelength of light,given by Cauchy's equation: $\mu = A + \frac{B}{\lambda^2}$.
Since the wavelength of red light $(\lambda_{red})$ is greater than the wavelength of blue light $(\lambda_{blue})$,the refractive index for red light is less than that for blue light $(\mu_{red} < \mu_{blue})$.
Therefore,the ratio of the refractive index of red light to blue light is $\frac{\mu_{red}}{\mu_{blue}} < 1$.
Thus,the ratio is less than unity.

(A) The relationship between the wavelength of light in a medium $(\lambda_{medium})$ and its wavelength in a vacuum $(\lambda_{vacuum})$ is given by the formula: $\lambda_{medium} = \frac{\lambda_{vacuum}}{\mu}$,where $\mu$ is the refractive index of the medium.
Given: $\lambda_{vacuum} = 6000\;\mathring A$ and $\mu = 1.5$.
Substituting the values into the formula:
$\lambda_{medium} = \frac{6000}{1.5} = 4000\;\mathring A$.
Therefore,the wavelength of light in glass is $4000\;\mathring A$.

(D) When light travels from one medium to another, the frequency of the light remains constant because it depends on the source of the light. However, the velocity $(v)$ and wavelength $(\lambda)$ of the light change according to the relation $v = f \lambda$, where $f$ is the frequency. Since the refractive index $(n)$ of the medium changes, the velocity changes as $v = c/n$, which subsequently causes the wavelength to change to maintain the constant frequency. Therefore, only wavelength and velocity change.

(C) The frequency of light remains constant during refraction.
We know that the speed of light $v = f \lambda$,where $f$ is the frequency and $\lambda$ is the wavelength.
Since $f$ is constant,$v \propto \lambda$.
Also,the refractive index $\mu$ is defined as $\mu = c / v$,which implies $v = c / \mu$.
Therefore,$\lambda \propto 1 / \mu$.
Let $\lambda_1$ be the wavelength in air $(\mu_1 = 1)$ and $\lambda_2$ be the wavelength in glass $(\mu_2 = \mu)$.
Then,$\lambda_1 / \lambda_2 = \mu_2 / \mu_1 = \mu / 1 = \mu : 1$.

(A) When light travels from a denser medium (refractive index $\mu_1$) to a rarer medium (refractive index $\mu_2$),where $\mu_2 < \mu_1$,the speed of light $v$ is given by $v = \frac{c}{\mu}$.
Since the refractive index $\mu$ decreases,the velocity $v$ of the light beam increases.
The frequency of light depends only on the source and remains constant during refraction.
The wavelength $\lambda$ is given by $\lambda = \frac{v}{f}$. Since $v$ increases and $f$ is constant,the wavelength $\lambda$ also increases.
Therefore,the correct statement is that its velocity increases.

(B) According to Cauchy's dispersion formula,the refractive index $\mu$ of a material is related to the wavelength $\lambda$ by the relation $\mu \approx A + \frac{B}{\lambda^2}$.
This shows that the refractive index $\mu$ is inversely related to the wavelength $\lambda$.
Since the wavelength of infrared light is greater than the wavelength of ultraviolet light $(\lambda_{\text{infrared}} > \lambda_{\text{ultraviolet}})$,the refractive index for infrared light will be less than the refractive index for ultraviolet light.
Therefore,$\mu_{\text{infrared}} < \mu_{\text{ultraviolet}}$.

(D) The refractive index $\mu$ is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$: $\mu = \frac{c}{v}$.
Given,$\mu = 2.0$ and $c = 3 \times 10^8 \ m/s$.
Rearranging the formula for velocity: $v = \frac{c}{\mu}$.
Substituting the values: $v = \frac{3 \times 10^8 \ m/s}{2.0} = 1.5 \times 10^8 \ m/s$.
To convert the velocity from $m/s$ to $cm/s$,we multiply by $100$ $(1 \ m = 100 \ cm)$:
$v = 1.5 \times 10^8 \times 10^2 \ cm/s = 1.5 \times 10^{10} \ cm/s$.

(B) According to Snell's Law,$n(A) \sin(i) = n(B) \sin(r)$.
Given that the angle of incidence $i$ is greater than the angle of refraction $r$ $(i > r)$,it follows that $\sin(i) > \sin(r)$.
For the equation to hold,we must have $n(A) < n(B)$,which means medium $B$ is optically denser than medium $A$.
The speed of light in a medium is given by $v = c/n$. Since $n(A) < n(B)$,it follows that $v(A) > v(B)$.
Therefore,the correct relationship is $v(A) > v(B)$ and $n(A) < n(B)$.

(C) The speed of light in a medium with refractive index $n$ is given by $v = c / n$,where $c$ is the speed of light in vacuum.
Time taken to travel a distance $t$ is given by the formula: $Time = \text{Distance} / \text{Speed}$.
Substituting the values,we get: $Time = t / (c / n) = (nt) / c$.
Therefore,the correct option is $C$.

(D) The time taken by light to travel a distance $d$ in a medium is given by $t = d/v$,where $v$ is the speed of light in that medium.
Since $v = c/\mu$,where $c$ is the speed of light in vacuum and $\mu$ is the refractive index,we have $t = d\mu/c$.
Let $t_a$ be the time taken in vacuum (or air) and $t_w$ be the time taken in water.
Then,$t_a = d/c$ and $t_w = d/v_w = d\mu_w/c$.
Therefore,$t_w = t_a \times \mu_w$.
Given $t_a = 8$ min $20$ sec $= 500$ sec and $\mu_w = 4/3$.
$t_w = 500 \times (4/3) = 2000/3$ sec.
$t_w = 666.67$ sec.
Converting to minutes: $666.67 / 60 = 11.11$ min.
$0.11$ min $= 0.11 \times 60 = 6.6$ sec $\approx 6$ sec.
Thus,$t_w = 11$ min $6$ sec.

(A) The optical path length is defined as the product of the refractive index of the medium and the geometric distance traveled by light in that medium.
Optical path = $\mu \times d$
For medium $1$,the optical path is $\mu_1 d_1$.
For medium $2$,the optical path is $\mu_2 d_2$.
Therefore,the total optical path length for the two media in contact is the sum of the individual optical paths:
Total optical path = $\mu_1 d_1 + \mu_2 d_2$.

(B) An object becomes invisible in a transparent medium if the refractive index of the object is equal to the refractive index of the medium. This is because there is no refraction or reflection of light at the interface between the object and the medium.
The refractive index of the glass piece is $\mu_{glass} = 1.61$.
From the given data:
- Refractive index of liquid $A = 1.51$
- Refractive index of liquid $B = 1.53$
- Refractive index of liquid $C = 1.61$
- Refractive index of liquid $D = 1.52$
- Refractive index of liquid $E = 1.65$
Since the refractive index of liquid $C$ is equal to the refractive index of the glass piece $(1.61 = 1.61)$,the glass piece will not be visible when it is in liquid $C$.

(B) Given:
Refractive index of glass with respect to air,$_a\mu_g = 3/2$
Refractive index of water with respect to air,$_a\mu_w = 4/3$
We need to find the refractive index of glass with respect to water,$_w\mu_g$.
The formula for relative refractive index is:
$_w\mu_g = \frac{_a\mu_g}{_a\mu_w}$
Substituting the given 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 $9/8$.

(C) The refractive index of medium $j$ with respect to medium $i$ is defined as $_i{\mu _j} = \frac{\mu_j}{\mu_i}$,where $\mu$ represents the absolute refractive index of the medium.
Given the expression: $_2{\mu _1} \times {\,_3}{\mu _2} \times {\,_4}{\mu _3}$.
Substituting the definition: $\frac{\mu_1}{\mu_2} \times \frac{\mu_2}{\mu_3} \times \frac{\mu_3}{\mu_4}$.
Canceling the common terms $\mu_2$ and $\mu_3$,we get: $\frac{\mu_1}{\mu_4}$.
By definition,$\frac{\mu_1}{\mu_4} = {_4}{\mu _1}$.
Since $_4{\mu _1} = \frac{1}{{_1{\mu _4}}}$,the product is equal to $\frac{1}{{_1{\mu _4}}}$.

(A) The colour of light is determined by its frequency.
When light travels from one medium to another,its speed and wavelength change,but its frequency remains constant.
Since the frequency of the light does not change,the perceived colour of the object remains the same.
Therefore,the diver sees the object as green.

(A) The refractive index of medium $2$ with respect to medium $1$ is defined as $_1{n_2} = \frac{n_2}{n_1}$.
For a series of media,the product of refractive indices is:
$_a{n_w} \times {_w}{n_{gl}} \times {_{gl}}{n_{gas}} \times {_{gas}}{n_a} = \left( \frac{n_w}{n_a} \right) \times \left( \frac{n_{gl}}{n_w} \right) \times \left( \frac{n_{gas}}{n_{gl}} \right) \times \left( \frac{n_a}{n_{gas}} \right) = 1$.
Thus,the product of the refractive indices for a closed loop of media returns to the initial medium is equal to $1$.

(B) The frequency of light remains constant when it travels from one medium to another. The speed of light $v$ is related to wavelength $\lambda$ and frequency $f$ by the equation $v = f \lambda$.
Since $f$ is constant,$v \propto \lambda$.
Therefore,$\frac{v_w}{v_a} = \frac{\lambda_w}{\lambda_a}$,where $v_w$ and $\lambda_w$ are the speed and wavelength in water,and $v_a$ and $\lambda_a$ are the speed and wavelength in air.
Given $v_a = 3 \times 10^8 \ m/s$,$\lambda_a = 6000 \ \mathring A$,and $\lambda_w = 4500 \ \mathring A$.
$v_w = v_a \times \frac{\lambda_w}{\lambda_a} = (3 \times 10^8 \ m/s) \times \frac{4500}{6000}$.
$v_w = 3 \times 10^8 \times 0.75 = 2.25 \times 10^8 \ m/s$.

(C) 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: $\lambda_a = 4200 \, \mathring{A}$ and $\mu = 4/3$.
Substituting the values:
$\lambda_m = \frac{4200}{4/3} = 4200 \times \frac{3}{4} = 1050 \times 3 = 3150 \, \mathring{A}$.
Therefore,the wavelength of the light in water is $3150 \, \mathring{A}$.

(C) The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in vacuum $(C)$ to the speed of light in that medium $(v)$:
$\mu = \frac{C}{v}$
Given that the refractive index $\mu = 1.5$ and the speed of light in vacuum is $C$,we can rearrange the formula to solve for the velocity in the medium $(v)$:
$v = \frac{C}{\mu} = \frac{C}{1.5}$
Therefore,the velocity of light in the medium is $\frac{C}{1.5}$.

(B) When light travels from one medium to another,its frequency remains constant because it depends on the source of light.
Since the refractive index of water $(n_w \approx 1.33)$ is greater than that of air $(n_a \approx 1.0)$,the speed of light decreases in water $(v = c/n)$.
According to the relation $v = f \lambda$,since $v$ decreases and $f$ remains constant,the wavelength $\lambda$ must also decrease.
Therefore,the frequency remains the same,but the wavelength is smaller in water than in air.

(B) According to the problem,the angle of incidence $i = 60^o$.
Since the reflected and refracted rays are mutually perpendicular,the angle between them is $90^o$.
From the geometry of the reflection and refraction,we have $i + 90^o + r = 180^o$,where $r$ is the angle of refraction.
Substituting $i = 60^o$,we get $60^o + 90^o + r = 180^o$,which gives $r = 30^o$.
Using Snell's Law,$\mu = \frac{\sin i}{\sin r} = \frac{\sin 60^o}{\sin 30^o}$.
Substituting the values,$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(B) The refractive index $\mu$ of a medium is inversely proportional to the speed of light $v$ in that medium,given by $\mu = c/v$,where $c$ is the speed of light in a vacuum.
Therefore,$\mu \propto 1/v$,which implies $\mu_g v_g = \mu_w v_w$.
Given: $\mu_g = 3/2$,$\mu_w = 4/3$,and $v_g = 2.00 \times 10^8 \ m/s$.
Using the relation $\frac{\mu_g}{\mu_w} = \frac{v_w}{v_g}$:
$\frac{3/2}{4/3} = \frac{v_w}{2.00 \times 10^8 \ m/s}$
$\frac{9}{8} = \frac{v_w}{2.00 \times 10^8 \ m/s}$
$v_w = \frac{9}{8} \times 2.00 \times 10^8 \ m/s = 2.25 \times 10^8 \ m/s$.

(A) The frequency of light remains unchanged when it enters a different medium. The frequency is $\nu = 5 \times 10^{14} \ Hz$.
The speed of light in vacuum is $c = 3 \times 10^8 \ m/s$.
The wavelength in vacuum is $\lambda_a = \frac{c}{\nu} = \frac{3 \times 10^8}{5 \times 10^{14}} = 0.6 \times 10^{-6} \ m = 6000 \ \mathring{A}$.
The wavelength in a medium of refractive index $\mu$ is given by $\lambda_m = \frac{\lambda_a}{\mu}$.
Given $\mu = 1.5$,we have $\lambda_m = \frac{6000 \ \mathring{A}}{1.5} = 4000 \ \mathring{A}$.

(B) The wavelength of light in a medium is given by the formula $\lambda_{medium} = \frac{\lambda_{air}}{\mu}$.
Given,$\lambda_{air} = 7200 \ \mathring{A}$ and refractive index $\mu = 1.5$.
Substituting the values,we get $\lambda_{glass} = \frac{7200}{1.5} = 4800 \ \mathring{A}$.
Therefore,the correct option is $B$.

(D) The refractive index of a medium $2$ with respect to medium $1$ is given by $_1\mu_2 = \frac{\mu_2}{\mu_1}$.
Given the expression in option $D$: $\frac{_a\mu_r}{_w\mu_r}$.
Substituting the refractive indices: $\frac{_a\mu_r}{_w\mu_r} = \frac{\mu_r / \mu_a}{\mu_r / \mu_w}$.
Simplifying the fraction: $\frac{\mu_r}{\mu_a} \times \frac{\mu_w}{\mu_r} = \frac{\mu_w}{\mu_a}$.
By definition,$\frac{\mu_w}{\mu_a} = _a\mu_w$.
Therefore,the relation $\frac{_a\mu_r}{_w\mu_r} = _a\mu_w$ is correct.

(D) The speed of light in a medium is given by $v = \frac{c}{\mu}$,where $c = 3 \times 10^8 \ m/s$ is the speed of light in vacuum and $\mu = 1.5$ is the refractive index of glass.
Substituting the values,we get $v = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \ m/s$.
The time given is $t = 1 \text{ nanosecond} = 10^{-9} \ s$.
The distance travelled is $d = v \times t = (2 \times 10^8 \ m/s) \times (10^{-9} \ s) = 0.2 \ m$.
Converting meters to centimeters,$d = 0.2 \times 100 \ cm = 20 \ cm$.

(C) When light travels from one medium to another,its frequency $(f)$ remains constant because it depends on the source of the light.
The speed of light $(v)$ in a medium is given by $v = f \lambda$,where $\lambda$ is the wavelength.
Since the refractive index of glass $(\mu_g)$ is greater than that of air $(\mu_a)$,the speed of light decreases when it enters the glass $(v_g < v_a)$.
Because $v = f \lambda$ and $f$ is constant,the decrease in speed $(v)$ must result in a decrease in wavelength $(\lambda)$.
Therefore,the wavelength decreases while the frequency remains unchanged.

(C) The real depth of the liquid is $d = 1 \ m$.
The apparent shift in the position of the mark is $\Delta d = 0.1 \ m$.
The apparent depth $d'$ is given by $d' = d - \Delta d = 1 - 0.1 = 0.9 \ m$.
The refractive index $\mu$ is defined as the ratio of real depth to apparent depth:
$\mu = \frac{d}{d'} = \frac{1}{0.9} = \frac{10}{9}$.

(A) When a liquid is heated,its density generally decreases because the molecules move further apart due to increased thermal energy.
According to the Lorentz-Lorenz formula,the refractive index $n$ is related to the density $\rho$ of the medium.
As the temperature increases,the density $\rho$ decreases,which leads to a decrease in the refractive index $n$.
Therefore,the refractive index of a liquid generally decreases upon heating.

(C) Snell's law in vector form relates the incident ray vector $\hat{i}$,the refracted ray vector $\hat{r}$,and the normal vector $\hat{n}$ to the interface.
According to the law of refraction,$\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2$.
In vector notation,where $\hat{n}$ is the normal directed towards the incident medium,the cross product $\hat{i} \times \hat{n}$ has a magnitude of $\sin \theta_1$ (since $\hat{i}$ and $\hat{n}$ are unit vectors).
Similarly,$\hat{r} \times \hat{n}$ has a magnitude of $\sin \theta_2$.
Thus,the vector form of Snell's law is expressed as $\hat{i} \times \hat{n} = \mu (\hat{r} \times \hat{n})$,where $\mu$ is the refractive index of the second medium relative to the first.

(A) When light travels from a denser medium (liquid) to a rarer medium (air),it bends away from the normal.
Due to this phenomenon of refraction,the light rays coming from the bottom of the container appear to originate from a point higher than their actual position.
This makes the bottom of the container appear slightly raised to an observer looking from above.
Therefore,the correct option is $A$.

(C) The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in vacuum (or air) $c$ to the speed of light in that medium $v$.
Formula: $\mu = \frac{c}{v}$
Given:
Speed of light in air,$c = 3 \times 10^8 \, m/s$
Refractive index of diamond,$\mu = 2.4$
Rearranging the formula to find the speed in diamond $(v)$:
$v = \frac{c}{\mu}$
$v = \frac{3 \times 10^8}{2.4}$
$v = 1.25 \times 10^8 \, m/s$
Therefore,the speed of light in diamond is $1.25 \times 10^8 \, m/s$.

(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 $(v)$: $\mu = \frac{c}{v}$.
Given: $\mu = 1.33$ and $c = 3 \times 10^8 \ m/s$.
Rearranging the formula to solve for $v$: $v = \frac{c}{\mu}$.
Substituting the values: $v = \frac{3 \times 10^8}{1.33} \approx 2.25 \times 10^8 \ m/s$.
Therefore,the speed of light in water is $2.25 \times 10^8 \ m/s$.

(B) Given:
Refractive index of water with respect to air,$^a\mu_w = 4/3$.
Refractive index of glass with respect to air,$^a\mu_g = 3/2$.
We need to find the refractive index of water with respect to glass,denoted as $^g\mu_w$.
The formula for the refractive index of medium $2$ with respect to medium $1$ is given by $^1\mu_2 = \frac{\mu_2}{\mu_1}$.
Therefore,$^g\mu_w = \frac{^a\mu_w}{^a\mu_g}$.
Substituting the given values:
$^g\mu_w = \frac{4/3}{3/2} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$.
Thus,the refractive index of water with respect to glass is $8/9$.

(B) When light travels from one medium to another,its frequency $(n)$ remains constant because it depends on the source of the light.
When light enters a medium with refractive index $\mu > 1$,its velocity $(v')$ decreases and is given by $v' = \frac{v}{\mu}$.
Since the relation between velocity,frequency,and wavelength is $v = n\lambda$,and $n$ is constant,the new wavelength $\lambda'$ is given by $\lambda' = \frac{v'}{n} = \frac{v/\mu}{n} = \frac{\lambda}{\mu}$.
Therefore,the new frequency,wavelength,and velocity are $n, \frac{\lambda}{\mu}, \frac{v}{\mu}$ respectively.

(B) According to Snell's Law,the refractive index $\mu$ of the medium with respect to air is given by $\mu = \frac{\sin i}{\sin r}$.
Also,the refractive index is related to the velocity of light in vacuum $(c)$ and in the medium $(v)$ by $\mu = \frac{c}{v}$.
Equating these,we get $\frac{c}{v} = \frac{\sin i}{\sin r}$.
Given $i = 45^\circ$,$r = 30^\circ$,and $c = 3 \times 10^8 \ m/s$.
Substituting the values: $\frac{3 \times 10^8}{v} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
Therefore,$v = \frac{3 \times 10^8}{\sqrt{2}} \approx \frac{3 \times 10^8}{1.414} \approx 2.12 \times 10^8 \ m/s$.

(C) The velocity of light $v$ in a medium is inversely proportional to its refractive index $\mu$,given by the relation $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum.
Therefore,the ratio of the velocity of light in glass $(v_g)$ to the velocity of light in water $(v_w)$ is given by:
$\frac{v_g}{v_w} = \frac{\mu_w}{\mu_g}$
Given $\mu_g = \frac{3}{2}$ and $\mu_w = \frac{4}{3}$.
Substituting these values:
$\frac{v_g}{v_w} = \frac{4/3}{3/2} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$
Thus,the ratio is $8:9$.

(B) The time taken $t$ by light to travel a distance $x$ through a medium of refractive index $\mu$ is given by $t = \frac{\mu x}{c}$,where $c$ is the speed of light in a vacuum.
Since the time taken is equal for both mediums,we have $t_A = t_B$.
Therefore,$\frac{\mu_A x_A}{c} = \frac{\mu_B x_B}{c}$.
This simplifies to $\frac{\mu_B}{\mu_A} = \frac{x_A}{x_B}$.
Given the ratio of thicknesses $x_A : x_B = 6 : 4$,we substitute these values:
$\frac{\mu_B}{\mu_A} = \frac{6}{4} = 1.5$.
The refractive index of $B$ with respect to $A$ is denoted as $_A\mu_B = \frac{\mu_B}{\mu_A} = 1.5$.

(D) Given that the refractive index of water with respect to air is $_a\mu_w = 1.3$.
Given that the refractive index of glass with respect to air is $_a\mu_g = 1.5$.
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:
$_w\mu_g = \frac{1.5}{1.3}$
Therefore,the correct option is $D$.

(A) The refractive index $\mu$ is defined as the ratio of real depth to apparent depth.
Given:
Real depth $d = 120 \,mm$
Apparent depth $d' = 80 \,mm$
Formula: $\mu = \frac{d}{d'}$
Calculation: $\mu = \frac{120}{80} = 1.5$
Therefore,the refractive index of benzene is $1.5$.

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

(B) The optical path length is defined as the product of the refractive index $(\mu)$ and the geometric path length $(x)$.
Given that the optical path is the same for both media,we have:
$\mu_{glass} \times x_{glass} = \mu_{water} \times x_{water}$
Substituting the given values:
$1.53 \times 4.0 = \mu_{water} \times 4.5$
$\mu_{water} = \frac{1.53 \times 4.0}{4.5}$
$\mu_{water} = \frac{6.12}{4.5} = 1.36$
Therefore,the refractive index of water is $1.36$.

(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 medium,the velocity of light is given by $v = c/n$,where $n$ is the refractive index of the medium.
Since the refractive index $n$ of any material medium is always greater than $1$,the velocity of light in any medium is always less than the velocity of light in a vacuum.
Therefore,the velocity of light is maximum in a vacuum.

(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 a vacuum.
For glass $(g)$ and liquid $(l)$:
$\mu_g = \frac{c}{v_g}$ and $\mu_l = \frac{c}{v_l}$
Taking the ratio:
$\frac{\mu_l}{\mu_g} = \frac{v_g}{v_l}$
Given:
$\mu_g = 1.5$
$v_g = 2 \times 10^8 \ m/s$
$v_l = 2.5 \times 10^8 \ m/s$
Substituting the values:
$\frac{\mu_l}{1.5} = \frac{2 \times 10^8}{2.5 \times 10^8}$
$\frac{\mu_l}{1.5} = \frac{2}{2.5} = 0.8$
$\mu_l = 1.5 \times 0.8 = 1.2$
Thus,the refractive index of the liquid is $1.2$.

(D) According to Snell's Law at the air-oil interface:
${\mu _{air}} \sin i = {\mu _{oil}} \sin {r_1}$
Given $i = 40^o$,${\mu _{air}} = 1$,${\mu _{oil}} = 1.45$:
$1 \times \sin 40^o = 1.45 \times \sin {r_1}$
$\sin {r_1} = \frac{\sin 40^o}{1.45} = \frac{0.6428}{1.45} \approx 0.4433$
Now,applying Snell's Law at the oil-water interface:
${\mu _{oil}} \sin {r_1} = {\mu _{water}} \sin r$
Since ${\mu _{oil}} \sin {r_1} = \sin i$ (from the first interface):
$\sin i = {\mu _{water}} \sin r$
$\sin 40^o = 1.33 \times \sin r$
$\sin r = \frac{\sin 40^o}{1.33} = \frac{0.6428}{1.33} \approx 0.4833$
$r = \arcsin(0.4833) \approx 28.9^o$

(D) When an object is immersed in a fluid,light rays undergo refraction at the interface of the object and the fluid.
If the refractive index of the object $(n_1)$ is equal to the refractive index of the surrounding fluid $(n_2)$,then the light rays do not bend or reflect at the interface.
As a result,the object becomes invisible because there is no change in the optical path of the light passing through the medium.
Therefore,the correct condition is that the refractive index of the object must exactly match that of the surrounding fluid.

(B) According to Snell's Law,$n_1 \sin \theta_1 = n_2 \sin \theta_2$.
Here,light travels from glass (denser medium,$n_1 \approx 1.5$) to air (rarer medium,$n_2 \approx 1.0$).
Since $n_1 > n_2$,the light ray bends away from the normal.
Therefore,the angle of refraction ${\theta _2}$ must be greater than the angle of incidence ${\theta _1}$.
Thus,the correct relation is ${\theta _1} < {\theta _2}$.

(D) The refractive index $\mu$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in the medium $(v)$.
Given:
Speed of light in vacuum,$c = 3 \times 10^8 \ m/s$
Speed of light in the medium,$v = 1.5 \times 10^8 \ m/s$
Formula:
$\mu = \frac{c}{v}$
Calculation:
$\mu = \frac{3 \times 10^8 \ m/s}{1.5 \times 10^8 \ m/s} = \frac{3}{1.5} = 2$
Therefore,the refractive index of the medium is $2$.