Refraction of Light Questions in English

(TOWARDS THE NORMAL) The light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium, it slows down and bends towards the normal. Since water $(n \approx 1.33)$ is optically denser than air $(n \approx 1.00)$, a ray of light travelling from air into water will bend towards the normal.

(D) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$:
$n = \frac{c}{v}$
Given:
Speed of light in vacuum $(c)$ = $3 \times 10^8 \, m \, s^{-1}$
Refractive index of glass $(n)$ = $1.50$
Rearranging the formula to find the speed of light in glass $(v)$:
$v = \frac{c}{n}$
$v = \frac{3 \times 10^8 \, m \, s^{-1}}{1.50}$
$v = 2 \times 10^8 \, m \, s^{-1}$
Therefore,the speed of light in glass is $2 \times 10^8 \, m \, s^{-1}$.

(N/A) The optical density of a medium is directly related to its refractive index. $A$ medium with a higher refractive index has a higher optical density,and a medium with a lower refractive index has a lower optical density.
$1$. Highest optical density: By comparing the refractive indices in the table,Diamond has the highest refractive index of $2.42$. Therefore,Diamond has the highest optical density.
$2$. Lowest optical density: By comparing the refractive indices in the table,Air has the lowest refractive index of $1.0003$. Therefore,Air has the lowest optical density.

(WATER) The speed of light in a medium is related to its refractive index $(n_m)$ by the formula:
$n_m = \frac{\text{Speed of light in vacuum (or air)}}{\text{Speed of light in the medium}} = \frac{c}{v}$
From this,the speed of light in the medium $(v)$ is given by:
$v = \frac{c}{n_m}$
This shows that the speed of light $(v)$ is inversely proportional to the refractive index $(n_m)$: $v \propto \frac{1}{n_m}$.
Therefore,light travels fastest in the medium with the lowest refractive index.
Comparing the given refractive indices:
- Water: $1.33$
- Kerosene: $1.44$
- Turpentine oil: $1.47$
Since water has the lowest refractive index $(1.33)$ among the three,light travels fastest in water.

(N/A) The refractive index of a medium $n_m$ is related to the speed of light in that medium $v$ by the relation:
$n_m = \frac{\text{Speed of light in air}}{\text{Speed of light in the medium}} = \frac{c}{v}$
Where,$c$ is the speed of light in vacuum or air.
The refractive index of diamond is $2.42$. This suggests that the speed of light in diamond is reduced by a factor of $2.42$ compared to its speed in air.

(D) According to Snell's law,the refractive index of medium $B$ relative to medium $A$ $(n_{BA})$ is given by the ratio of the sine of the angle of incidence $(i)$ to the sine of the angle of refraction $(r)$:
$n_{BA} = \frac{\sin i}{\sin r}$
From the figure,the light ray travels from medium $A$ to medium $B$. The angle of incidence $i$ is the angle between the incident ray and the normal in medium $A$. Given the angle with the normal is $60^\circ$,so $i = 60^\circ$.
The angle of refraction $r$ is the angle between the refracted ray and the normal in medium $B$. Given the angle with the normal is $45^\circ$,so $r = 45^\circ$.
Substituting these values into Snell's law:
$n_{BA} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \times \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$
Therefore,the refractive index of medium $B$ relative to medium $A$ is $\sqrt{3} / \sqrt{2}$.

(A) From the figure,we can observe the angles of incidence $(i)$ and refraction $(r)$.
By comparing the angles with the normal,we see that the angle of refraction $(r)$ is smaller than the angle of incidence $(i)$,i.e.,$r < i$.
According to Snell's Law,the refractive index of medium $B$ with respect to medium $A$ is given by $n_{BA} = \frac{\sin i}{\sin r}$.
Since $i > r$,it follows that $\sin i > \sin r$,which implies that $\frac{\sin i}{\sin r} > 1$.
Therefore,the refractive index of medium $B$ relative to $A$ is greater than unity.

(B) By observing the figure,we can see that the incident light beams entering through holes $A$ and $B$ emerge through holes $C$ and $D$ respectively,without any change in their direction or convergence/divergence.
This indicates that the light rays have undergone lateral displacement,which is a characteristic property of a rectangular glass slab.
$A$ convex lens would converge the rays,a concave lens would diverge them,and a prism would deviate the rays at an angle.
Therefore,a rectangular glass slab is the correct component inside the box.

(B) When a light ray travels from a rarer medium (air) to a denser medium (glass),it bends towards the normal at the point of incidence.
When the light ray emerges from the denser medium (glass) back into the rarer medium (air),it bends away from the normal.
In figure $B$,the ray bends towards the normal upon entering the glass slab and bends away from the normal upon exiting it,which correctly follows the laws of refraction.
Therefore,the path shown in $B$ is correct.

(C) The bending of light (refraction) depends on the refractive index of the medium. $A$ higher refractive index means the medium is optically denser,causing light to bend more towards the normal when entering from air.
The refractive indices of the given media are approximately:
$1$. Water: $1.33$
$2$. Kerosene: $1.44$
$3$. Mustard oil: $1.46$
$4$. Glycerine: $1.47$
Since glycerine has the highest refractive index among the given options,a ray of light incident obliquely at the same angle will bend the most in glycerine.

(N/A) When a light ray enters a rectangular glass slab,it undergoes refraction at two parallel interfaces.
$1$. At the first interface (air-glass),the light ray bends towards the normal because it travels from a rarer to a denser medium.
$2$. At the second interface (glass-air),the light ray bends away from the normal as it travels from a denser to a rarer medium.
$3$. According to Snell's Law,the angle of incidence at the first surface is equal to the angle of emergence at the second surface because the opposite sides of the rectangular slab are parallel.
$4$. Consequently,the emergent ray is parallel to the incident ray,but it undergoes a lateral displacement.

(B) No,the pencil will not appear to be bent to the same extent. The extent of bending of the pencil depends on the refractive index of the liquid used. Since different liquids like kerosene and turpentine have different refractive indices compared to water,the angle of refraction will vary. The bending of light at the interface of two media is determined by the relative refractive index,which dictates the change in the velocity of light as it passes from one medium to another.

(N/A) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$. Mathematically,it is expressed as: $n = \frac{c}{v}$.
To find the refractive index of medium $2$ with respect to medium $1$ $(n_{21})$,we take the ratio of the speed of light in medium $1$ $(v_1)$ to the speed of light in medium $2$ $(v_2)$. The expression is: $n_{21} = \frac{v_1}{v_2}$.

(D) The refractive index of diamond with respect to glass is given by $n_{dg} = \frac{n_d}{n_g} = 1.6$.
Here,$n_d$ is the absolute refractive index of diamond and $n_g$ is the absolute refractive index of glass.
Given that $n_g = 1.5$.
Substituting the values,we get $\frac{n_d}{1.5} = 1.6$.
Therefore,$n_d = 1.6 \times 1.5 = 2.4$.

(N/A) When light travels from a rarer medium (air) to a denser medium (water),it bends towards the normal. In this case,the angle of refraction $(r)$ is less than the angle of incidence $(i)$.
When light travels from a denser medium (water) to a rarer medium (air),it bends away from the normal. In this case,the angle of refraction $(r)$ is greater than the angle of incidence $(i)$.
The ray diagrams are provided in the image.

(N/A) The laws of refraction are as follows:
$1$. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
$2$. Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given color and for a given pair of media, expressed as $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$.
When a light ray passes through a rectangular glass slab:
- The ray enters from air to glass (denser medium) and bends towards the normal.
- It travels through the glass and exits from glass to air (rarer medium), bending away from the normal.
- The emergent ray is parallel to the incident ray but laterally displaced. This perpendicular distance between the original path of the incident ray and the emergent ray is called lateral displacement.

(C) The refractive index $(n)$ is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in a given medium $(v)$,expressed as $n = c/v$.
Since it is a ratio of two similar physical quantities (both having units of speed,$m/s$),the units cancel out.
Therefore,the refractive index is a dimensionless quantity and has no units.

(B) When light travels from a rarer medium to a denser medium, its speed decreases.
Since the frequency $(f)$ of light remains constant during refraction, the relationship between speed $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by $v = f \lambda$.
Because $v$ decreases and $f$ is constant, the wavelength $(\lambda)$ must also decrease.

(N/A) Snell's law states that for a given pair of media,the ratio of the sine of the angle of incidence $(i)$ to the sine of the angle of refraction $(r)$ is constant.
Mathematically,it is expressed as: $\frac{\sin i}{\sin r} = n_{21}$,where $n_{21}$ is the refractive index of the second medium with respect to the first medium.
This constant value is known as the refractive index of the second medium with respect to the first.

(N/A) When a ray of light passes through a rectangular glass slab,it undergoes refraction at two parallel surfaces. The emergent ray is parallel to the incident ray but is displaced sideways from the original path of the incident ray. This perpendicular distance between the original path of the incident ray and the emergent ray is known as the lateral shift.

(A) The velocity of light in a medium is calculated using the formula $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum $(3 \times 10^{8} \ m \ s^{-1})$ and $\mu$ is the refractive index of the medium.
Given,$\mu = 1.5$.
Substituting the values: $v = \frac{3 \times 10^{8}}{1.5} = 2 \times 10^{8} \ m \ s^{-1}$.
Therefore,the velocity of light in the glass slab is $2 \times 10^{8} \ m \ s^{-1}$.

(A) According to Snell's Law,the degree of bending (refraction) of light depends on the refractive index of the medium for that specific wavelength.
Light that bends more has a higher refractive index for that medium,which implies a lower velocity of light in that medium.
Since colour $A$ is bent more than colour $B$,the refractive index for colour $A$ is higher than that for colour $B$ in the second medium.
Because the speed of light $v$ is inversely proportional to the refractive index $n$ $(v = c/n)$,a higher refractive index results in a lower speed.
Therefore,colour $A$ travels more slowly in the second medium.

(B) When light travels from one medium to another, its frequency remains constant because it depends on the source of the light. However, the speed of light changes as it enters a different medium. Since the relationship between speed $(v)$, frequency $(f)$, and wavelength $(\lambda)$ is given by $v = f \times \lambda$, if the speed changes and the frequency remains constant, the wavelength must change. Therefore, when light passes from a rarer to a denser medium, its wavelength decreases.

(A) When a light ray falls normally (perpendicularly) on the surface of a glass slab,the angle of incidence $(i)$ is $0^{\circ}$.
According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$.
Since $\sin(0^{\circ}) = 0$,it follows that $\sin(r) = 0$,which means the angle of refraction $(r)$ is $0^{\circ}$.
Therefore,the ray passes through the glass slab without any deviation.

(N/A) Given: Refractive index of glass $(n_g)$ = $3/2$,Refractive index of water $(n_w)$ = $4/3$,Speed of light in glass $(v_g)$ = $2 \times 10^{8} \ m \ s^{-1}$.
$(i)$ Speed of light in vacuum $(c)$:
Using the formula $n_g = c / v_g$,we get $c = n_g \times v_g$.
$c = (3/2) \times (2 \times 10^{8} \ m \ s^{-1}) = 3 \times 10^{8} \ m \ s^{-1}$.
$(ii)$ Speed of light in water $(v_w)$:
Using the formula $n_w = c / v_w$,we get $v_w = c / n_w$.
$v_w = (3 \times 10^{8} \ m \ s^{-1}) / (4/3) = (3 \times 3 \times 10^{8}) / 4 = 2.25 \times 10^{8} \ m \ s^{-1}$.

(B) Snell's law is given by the formula $n_1 \sin(i) = n_2 \sin(r)$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
When light is incident normally (perpendicularly) on the surface,the angle of incidence $i = 0^\circ$.
Since $\sin(0^\circ) = 0$,the equation becomes $n_1(0) = n_2 \sin(r)$,which implies $\sin(r) = 0$,so $r = 0^\circ$.
In this case,the law does not provide a unique relationship between the refractive indices and the angles because the ratio $\frac{\sin(i)}{\sin(r)}$ becomes $\frac{0}{0}$,which is undefined.

(N/A) When a ray of light travels from one medium to another having equal refractive indices,no refraction or bending occurs. The light continues to travel in a straight line because there is no change in the optical density between the two media.
$(b)$ The cause of refraction of light is the change in the speed of light as it travels from one medium to another. When light enters a medium with a different optical density,its velocity changes,causing it to deviate from its original path.

(C) We know that the refractive index $n$ is given by $n = \frac{\sin i}{\sin r}$.
For a constant angle of incidence $i$,the refractive index $n$ is inversely proportional to $\sin r$.
Since the angle of refraction $r$ is minimum for medium $R$ $(15^{\circ})$,the refractive index $n$ is maximum for medium $R$.
We also know that the velocity of light $v$ in a medium is given by $v = \frac{c}{n}$,where $c$ is the speed of light in a vacuum.
Since $v$ is inversely proportional to $n$,the velocity of light will be minimum where the refractive index $n$ is maximum.
Therefore,the velocity of light is minimum in medium $R$.

(N/A) The absolute refractive index of a medium is defined as the ratio of the velocity of light in a vacuum to the velocity of light in that medium. If $c$ is the velocity of light in a vacuum and $v$ is the velocity in the medium,then $n = c / v$.
If $n_{a}$ and $n_{b}$ are the absolute refractive indices of media $A$ and $B$ respectively,then $n_{a} = c / v_{a}$ and $n_{b} = c / v_{b}$.
The refractive index of medium $B$ with respect to medium $A$ $(_{a}n_{b})$ is given by:
$_{a}n_{b} = \frac{\text{Velocity of light in } A}{\text{Velocity of light in } B} = \frac{v_{a}}{v_{b}} = \frac{c/n_{a}}{c/n_{b}} = \frac{n_{b}}{n_{a}}$.
Regarding the velocity of light: The velocity of light is inversely proportional to the optical density of the medium. As the optical density of a medium increases,the velocity of light in that medium decreases,and vice versa.

(A) Given: Refractive index of water $(n)$ = $4/3 = 1.33$.
Speed of light in vacuum $(c)$ = $3 \times 10^{8} \text{ m s}^{-1}$.
We know the formula for refractive index is $n = c / v$,where $v$ is the speed of light in the medium.
Rearranging the formula to find $v$: $v = c / n$.
Substituting the values: $v = (3 \times 10^{8}) / (4/3) = (3 \times 10^{8} \times 3) / 4 = 9 \times 10^{8} / 4 = 2.25 \times 10^{8} \text{ m s}^{-1}$.
Therefore,the speed of light in water is $2.25 \times 10^{8} \text{ m s}^{-1}$.

(N/A) In this case,the light ray travels from alcohol $(n=1.36)$ to kerosene $(n=1.44)$ and then to turpentine oil $(n=1.47)$. Since the light is moving from a rarer to a denser medium at each interface,the ray of light bends towards the normal.
$(b)$ In this case,the light ray travels from turpentine oil $(n=1.47)$ to kerosene $(n=1.44)$ and then to alcohol $(n=1.36)$. Since the light is moving from a denser to a rarer medium at each interface,the ray of light bends away from the normal.

(N/A) $(i)$ The incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane.
$(ii)$ The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media. This law is also known as Snell's law. Mathematically, $\frac{\sin i}{\sin r} = \text{constant} = {_1n}_2$.

(N/A) The ray diagram shows the path of a light ray passing through a rectangular glass slab. When a light ray enters the glass slab from air,it bends towards the normal because it travels from a rarer medium to a denser medium. Upon emerging from the glass slab into the air,it bends away from the normal.
$1$. Angle of incidence $(i)$: The angle between the incident ray and the normal at the point of incidence.
$2$. Angle of refraction $(r)$: The angle between the refracted ray and the normal inside the glass slab.
$3$. Angle of emergence $(e)$: The angle between the emergent ray and the normal at the second surface.
$4$. Lateral displacement: The perpendicular distance between the original path of the incident ray and the emergent ray.
The refractive index $(n)$ of the glass slab with respect to air is given by Snell's Law at the first interface:
$n = \frac{\sin i}{\sin r}$

(N/A) According to Snell's Law,$n_1 \sin(i) = n_2 \sin(r)$. For a constant angle of incidence $i$,the refractive index $n$ is inversely proportional to $\sin(r)$. Since the angle of refraction $r$ is smaller in medium $I$ $(20^{\circ})$ compared to medium $II$ $(30^{\circ})$,medium $I$ has a higher refractive index and is therefore optically denser.
$(b)$ The refractive index $n$ is given by the formula $n = \frac{c}{v}$,where $c$ is the speed of light in air (or vacuum) and $v$ is the speed of light in the medium.
Given $n = 2.42$ and $c = 3.00 \times 10^{8} \text{ m s}^{-1}$.
$v = \frac{c}{n} = \frac{3.00 \times 10^{8}}{2.42} \approx 1.24 \times 10^{8} \text{ m s}^{-1}$.

(N/A) When a ray of light falls normally (perpendicularly) on the surface of a glass slab,the angle between the incident ray and the normal is $0^{\circ}$. Therefore,the angle of incidence is $0^{\circ}$. Since the ray travels along the normal,it does not deviate,and the angle of refraction is also $0^{\circ}$.
$(b)$ Given:
Speed of light in air $(c)$ = $3 \times 10^{8} \ m \ s^{-1}$
Speed of light in medium $X$ $(v)$ = $1.5 \times 10^{8} \ m \ s^{-1}$
The refractive index $(n)$ of a medium is given by the ratio of the speed of light in air to the speed of light in that medium:
$n = \frac{c}{v}$
$n = \frac{3 \times 10^{8} \ m \ s^{-1}}{1.5 \times 10^{8} \ m \ s^{-1}}$
$n = 2$
Thus,the refractive index of medium $X$ is $2$.

(N/A) Alcohol is optically denser than water. This is because a higher refractive index indicates a higher optical density.
The ray diagram shows light bending towards the normal when passing from a rarer medium (water) to a denser medium (alcohol).
$(b)$ Given:
Absolute refractive index of diamond,$n_d = 2.42$
Absolute refractive index of glass,$n_g = 1.50$
Refractive index of diamond with respect to glass,$_g n_d = \frac{n_d}{n_g}$
$_g n_d = \frac{2.42}{1.50} = 1.6133$ (approximately $1.61$)

(N/A) Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media and for light of a given wavelength. Mathematically, $\frac{\sin i}{\sin r} = \text{constant} = n_{21}$, where $n_{21}$ is the refractive index of the second medium with respect to the first.
$(b)$ When a light ray enters a glass slab obliquely, it bends towards the normal as it travels from a rarer medium (air) to a denser medium (glass). Upon exiting the glass slab, it bends away from the normal as it travels from a denser medium to a rarer medium. The emergent ray is parallel to the incident ray but undergoes a lateral displacement.

(N/A) The two laws of refraction are:
$(i)$ The incident ray,the refracted ray,and the normal to the interface of two transparent media at the point of incidence,all lie in the same plane.
$(ii)$ The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant,for the light of a given colour and for the given pair of media. This law is also known as Snell's law of refraction.
$(b)$ Since the refractive index of medium $A$ $(n_A = 2.42)$ is greater than the refractive index of medium $B$ $(n_B = 1.65)$,the light travels from a denser medium to a rarer medium. Therefore,the refracted ray will bend away from the normal when entering medium $B$. When the ray emerges from the slab back into medium $A$,it will bend towards the normal. The path of the ray is as shown in the figure.

(A) Given: Refractive index of water with respect to air,$_{a}\mu_{w} = \frac{4}{3}$.
Refractive index of glass with respect to air,$_{a}\mu_{g} = \frac{3}{2}$.
We need to find the refractive index of glass with respect to water,$_{w}\mu_{g}$.
The formula is $_{w}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{w}}$.
Substituting the values: $_{w}\mu_{g} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$.

(A) Given:
Refractive index of dense flint glass with respect to air,$_{a}\mu_{g} = 1.65$
Refractive index of alcohol with respect to air,$_{a}\mu_{alc} = 1.36$
To find the refractive index of dense flint glass with respect to alcohol $(_{alc}\mu_{g})$,we use the formula:
$_{alc}\mu_{g} = \frac{_{a}\mu_{g}}{_{a}\mu_{alc}}$
Substituting the given values:
$_{alc}\mu_{g} = \frac{1.65}{1.36} \approx 1.21$
Therefore,the refractive index of dense flint glass with respect to alcohol is $1.21$.

(1.18) The refractive index of a medium $2$ with respect to medium $1$ is given by the formula: ${}_{1}n_{2} = \frac{{}_{a}n_{2}}{{}_{a}n_{1}}$.
Here,medium $1$ is ice $(i)$ and medium $2$ is rock salt $(r)$.
Given: Refractive index of ice with respect to air $({}_{a}n_{i})$ = $1.31$.
Refractive index of rock salt with respect to air $({}_{a}n_{r})$ = $1.54$.
Therefore,the refractive index of rock salt with respect to ice $({}_{i}n_{r})$ = $\frac{1.54}{1.31} \approx 1.18$.

(A) Given:
Speed of light in air $(c)$ = $3.0 \times 10^{8} \text{ m s}^{-1}$
Refractive index of diamond $(n)$ = $2.42$
Speed of light in diamond $(v)$ = ?
Formula:
The refractive index $(n)$ is defined as the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$:
$n = \frac{c}{v}$
Calculation:
Rearranging the formula to solve for $v$:
$v = \frac{c}{n}$
$v = \frac{3.0 \times 10^{8}}{2.42}$
$v \approx 1.24 \times 10^{8} \text{ m s}^{-1}$
Therefore,the speed of light in diamond is $1.24 \times 10^{8} \text{ m s}^{-1}$.

(D) When a light ray passes through a rectangular glass slab,it undergoes refraction at two parallel surfaces ($AB$ and $CD$).
According to the laws of refraction,when a light ray travels from a rarer medium (air) to a denser medium (glass),it bends towards the normal.
When it emerges from the denser medium (glass) back into the rarer medium (air),it bends away from the normal.
Because the opposite surfaces of the rectangular glass slab are parallel to each other,the emergent ray is parallel to the incident ray.
In the given diagram,the ray $S$ is parallel to the incident ray $IM$,which is the characteristic property of refraction through a rectangular glass slab.

(N/A) $(i)$ Light speeds up when it travels from an optically denser to an optically rarer medium. In other words, it travels from a medium of high refractive index to a medium of low refractive index. Therefore, one such pair is from crown glass $(1.52)$ into water $(1.33)$.
$(ii)$ Light slows down when it travels from an optically rarer to an optically denser medium. In other words, it travels from a medium of low refractive index to a medium of high refractive index. One such pair is from water $(1.33)$ into diamond $(2.42)$.

(WATER) The refractive index $(n)$ of a medium is defined by the formula $n = \frac{c}{v}$,where $c$ is the speed of light in a vacuum and $v$ is the speed of light in the given medium.
Since $c$ is a constant,the relationship between speed and refractive index is $v = \frac{c}{n}$,which implies $v \propto \frac{1}{n}$.
This means that the speed of light is inversely proportional to the refractive index of the medium.
Comparing the given refractive indices: Water $(1.33)$ < Kerosene $(1.44)$ < Turpentine $(1.47)$.
Since water has the lowest refractive index among the three,light travels the fastest through water.

(N/A) The lateral displacement depends upon:
$(i)$ The angle of incidence of the incident ray $PQ$ on the face $DC$ of the given glass slab.
$(ii)$ The thickness of the glass slab.
The lateral displacement is defined as the perpendicular distance between the emergent ray and the original path of the incident ray $PQ$ when produced forward.

(B) Optical density is closely related to the refractive index of a material. $A$ higher refractive index indicates a higher optical density.
The refractive indices of the given materials are approximately:
$1$. Water: $1.33$
$2$. Glass: $1.50$
$3$. Pearl: $1.53$
$4$. Diamond: $2.42$
Since diamond has the highest refractive index among the given options,it possesses the maximum optical density.

(B) The absolute refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$: $n = c/v$.
Since the speed of light in a vacuum is the maximum possible speed of light in any medium, $c$ is always greater than $v$ $(c > v)$.
Therefore, the ratio $c/v$ is always greater than $1$ for any medium other than a vacuum (where it is exactly $1$).
Thus, the absolute refractive index of any medium is always greater than $1$ $(n > 1)$.

(A) The relative refractive index of medium $2$ with respect to medium $1$ is given by the formula: $n_{21} = n_2 / n_1$, where $n_2$ is the refractive index of the second medium and $n_1$ is the refractive index of the first medium.
Given refractive indices: $n_{\text{water}} = 1.33$, $n_{\text{benzene}} = 1.50$, $n_{\text{sapphire}} = 1.77$.
Calculating the values for each option:
$A$: Sapphire relative to water = $1.77 / 1.33 \approx 1.3308$
$B$: Sapphire relative to benzene = $1.77 / 1.50 = 1.18$
$C$: Benzene relative to water = $1.50 / 1.33 \approx 1.1278$
$D$: Water relative to benzene = $1.33 / 1.50 \approx 0.8867$
Comparing these values, the relative refractive index of sapphire with respect to water $(1.3308)$ is the maximum.

(C) The refractive index $(n)$ of a medium is defined as the ratio of the speed of light in a vacuum $(c)$ to the speed of light in that medium $(v)$,given by $n = \frac{c}{v}$.
From this,the velocity of light in a medium is $v = \frac{c}{n}$.
Let $n_w = \frac{4}{3}$ be the refractive index of water and $n_g = \frac{3}{2}$ be the refractive index of glass.
The velocity of light in water is $v_w = \frac{c}{n_w}$ and in glass is $v_g = \frac{c}{n_g}$.
The ratio of the velocity of light in water to that in glass is $\frac{v_w}{v_g} = \frac{c/n_w}{c/n_g} = \frac{n_g}{n_w}$.
Substituting the given values: $\frac{v_w}{v_g} = \frac{3/2}{4/3} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8}$.
Therefore,the ratio is $\frac{9}{8}$.