Refraction through Plane Surface and Glass Slab Questions in English

(B) Let the actual distance of the bubble from one side be $x$. Then the distance from the opposite side is $(24 - x)$.
Using the formula for apparent depth: $d_{\text{apparent}} = \frac{d_{\text{actual}}}{\mu}$.
For the first side: $12 = \frac{x}{\mu} \implies x = 12\mu$.
For the second side: $4 = \frac{24 - x}{\mu} \implies 24 - x = 4\mu$.
Substituting $x = 12\mu$ into the second equation:
$24 - 12\mu = 4\mu$.
$24 = 16\mu$.
$\mu = \frac{24}{16} = 1.5 = \frac{3}{2}$.

(A) The apparent depth $d_{app}$ of a medium of real depth $d_i$ and refractive index $n_i$ is given by $d_{app, i} = \frac{d_i}{n_i}$.
Here,the vessel has a total depth $d$. It is divided into two halves,so the real depth of each liquid is $d_1 = d_2 = \frac{d}{2}$.
The apparent depth of the oil layer is $d_{app, 1} = \frac{d/2}{n_1} = \frac{d}{2n_1}$.
The apparent depth of the water layer is $d_{app, 2} = \frac{d/2}{n_2} = \frac{d}{2n_2}$.
The total apparent depth of the vessel is the sum of the apparent depths of the two layers:
$d_{app} = d_{app, 1} + d_{app, 2} = \frac{d}{2n_1} + \frac{d}{2n_2}$.
Factoring out $\frac{d}{2}$,we get $d_{app} = \frac{d}{2} \left( \frac{1}{n_1} + \frac{1}{n_2} \right)$.
Simplifying the expression inside the parentheses,we get $d_{app} = \frac{d}{2} \left( \frac{n_2 + n_1}{n_1 n_2} \right) = \frac{d(n_1 + n_2)}{2n_1 n_2}$.

(C) Let the velocity of the fish be $v_f = 8\,ms^{-1}$ (upward) and the velocity of the bird be $v_b$ (downward).
The apparent velocity of the bird as seen by the fish is given by $v_{b/f} = 12\,ms^{-1}$ (downward).
According to the formula for apparent velocity due to refraction at a plane surface,the apparent velocity of an object in a rarer medium as seen from a denser medium is $v_{app} = \mu \cdot v_{actual}$.
Here,the bird is in air $(\mu_1 = 1)$ and the fish is in water $(\mu_2 = 4/3)$.
The velocity of the bird relative to the water surface is $v_b$. The velocity of the fish relative to the water surface is $v_f = 8\,ms^{-1}$.
The apparent velocity of the bird relative to the fish is $v_{b/f} = v_{b,app} - v_f$.
Since the bird is in air,its apparent velocity as seen from water is $v_{b,app} = \mu \cdot v_b = \frac{4}{3} v_b$.
Given $v_{b/f} = 12\,ms^{-1}$ (downward),we have:
$12 = \frac{4}{3} v_b + 8$
$12 - 8 = \frac{4}{3} v_b$
$4 = \frac{4}{3} v_b$
$v_b = 3\,ms^{-1}$.

(D) The apparent depth of an object in a medium of refractive index $\mu$ and real depth $h$ is given by $h_{app} = \frac{h}{\mu}$.
For multiple layers of immiscible liquids,the total apparent depth is the sum of the apparent depths of each layer:
$h_{app} = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$
Given:
$h_1 = 6 \,cm, \mu_1 = \frac{8}{5}$
$h_2 = 6 \,cm, \mu_2 = \frac{3}{2}$
Substituting the values:
$h_{app} = \frac{6}{8/5} + \frac{6}{3/2}$
$h_{app} = \frac{30}{8} + \frac{12}{3}$
$h_{app} = \frac{15}{4} + 4 = \frac{15 + 16}{4} = \frac{31}{4} \,cm$
Comparing this with the given apparent depth $\frac{\alpha}{4} \,cm$,we get $\alpha = 31$.

(B) Given:
Thickness $t = 4 \sqrt{3} \text{ cm}$
Refractive index $\mu = \sqrt{2}$
Angle of incidence $i = \theta_c$ (critical angle)
$1$. Calculate the critical angle:
$\sin \theta_c = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$
$\theta_c = 45^{\circ}$
So,$i = 45^{\circ}$.
$2$. Apply Snell's Law at the first surface:
$1 \cdot \sin i = \mu \cdot \sin r$
$\sin 45^{\circ} = \sqrt{2} \cdot \sin r$
$\frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r$
$\sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}$.
$3$. Calculate lateral displacement $\Delta$:
The formula for lateral displacement is $\Delta = \frac{t \sin(i - r)}{\cos r}$.
$\Delta = \frac{4 \sqrt{3} \cdot \sin(45^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$
$\Delta = \frac{4 \sqrt{3} \cdot \sin 15^{\circ}}{\cos 30^{\circ}}$
Given $\sin 15^{\circ} = 0.25 = \frac{1}{4}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.
$\Delta = \frac{4 \sqrt{3} \cdot (1/4)}{\sqrt{3}/2} = \frac{\sqrt{3}}{\sqrt{3}/2} = 2 \text{ cm}$.
Thus,the lateral displacement is $2 \text{ cm}$.

(C) Let $h$ be the height of the ball above the water surface. The velocity of the ball at height $h$ is given by $v_b = \sqrt{2g(H - h)}$, where $H = 20 \,m$ is the initial height and $h = 12.8 \,m$.
$v_b = \sqrt{2 \times 10 \times (20 - 12.8)} = \sqrt{20 \times 7.2} = \sqrt{144} = 12 \,m/s$.
When an object is in a rarer medium (air) and viewed from a denser medium (water), the apparent height $h'$ is given by $h' = \mu h$, where $\mu = 4/3$ is the refractive index of water.
Therefore, the apparent velocity $v_a$ as seen by the fish is $v_a = \frac{d}{dt}(h') = \mu \frac{dh}{dt} = \mu v_b$.
$v_a = (4/3) \times 12 \,m/s = 16 \,m/s$.

(C) The optical path length of the light wave at the top of the slab is $n_2 d$,and at the bottom,it is $n_1 d$.
Since the refractive index varies linearly,the wavefront tilts as it passes through the slab.
The difference in optical path length between the top and bottom rays is $\Delta = (n_2 - n_1)d$.
This path difference over the height $h$ causes the wavefront to tilt by an angle $\theta$,where $\tan \theta = \frac{\Delta}{h} = \frac{(n_2 - n_1)d}{h}$.
Thus,the light wave deflects upwards by an angle $\theta = \tan^{-1}\left[\frac{(n_2 - n_1)d}{h}\right]$.
This confirms that statement $B$ is correct.
Since the deflection angle depends only on the difference $(n_2 - n_1)$,statement $D$ is also correct.
Therefore,the correct options are $B$ and $D$.

(C) The lateral displacement $x$ for one unit is given by the sum of horizontal shifts in each layer:
$x = h \tan 60^{\circ} + d \tan \theta_1 + d \tan \theta_2$
Using Snell's Law at each interface:
$1$. At the air-first layer interface: $1 \cdot \sin 60^{\circ} = \sqrt{\frac{3}{2}} \cdot \sin \theta_1 \Rightarrow \sin \theta_1 = \frac{\sqrt{3}/2}{\sqrt{3/2}} = \frac{1}{\sqrt{2}} \Rightarrow \theta_1 = 45^{\circ}$
$2$. At the first layer-second layer interface: $\sqrt{\frac{3}{2}} \cdot \sin 45^{\circ} = \sqrt{3} \cdot \sin \theta_2 \Rightarrow \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \sqrt{3} \cdot \sin \theta_2 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin \theta_2 \Rightarrow \sin \theta_2 = \frac{1}{2} \Rightarrow \theta_2 = 30^{\circ}$
Now,calculate the total horizontal shift $x$ for one unit:
$x = \frac{1}{3} \tan 60^{\circ} + \left(\frac{\sqrt{3}-1}{2}\right) \tan 45^{\circ} + \left(\frac{\sqrt{3}-1}{2}\right) \tan 30^{\circ}$
$x = \frac{1}{3} \cdot \sqrt{3} + \frac{\sqrt{3}-1}{2} \cdot 1 + \frac{\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3}}$
$x = \frac{\sqrt{3}}{3} + \frac{\sqrt{3}-1}{2} + \frac{3-\sqrt{3}}{6} = \frac{2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3}}{6} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}} \text{ cm}$
For $n$ units,the total distance $l = n \cdot x = n \cdot \frac{2}{\sqrt{3}} = \frac{8}{\sqrt{3}} \text{ cm}$
$n = 4$

(D) When a light ray passes through a parallel-sided glass slab of thickness $h$,it undergoes refraction at two parallel surfaces.
Let $i$ be the angle of incidence and $r$ be the angle of refraction.
The lateral shift $d$ is the perpendicular distance between the incident ray and the emergent ray.
From the geometry of the path of light through the slab,the lateral shift is given by the formula:
$d = \frac{h \sin(i - r)}{\cos r}$
Thus,the correct option is $D$.

(C) The apparent depth $d'$ of the bottom of the container when viewed from above through multiple layers of liquids is given by the formula:
$d' = \frac{h_1}{\mu_1} + \frac{h_2}{\mu_2}$
Here,$h_1 = 60 \ cm$,$\mu_1 = 1.2$,$h_2 = H$,and $\mu_2 = 1.6$.
So,$d' = \frac{60}{1.2} + \frac{H}{1.6} = 50 + \frac{H}{1.6}$.
The actual depth of the bottom is $d = 60 + H$.
The apparent shift is given by the difference between the actual depth and the apparent depth:
$\text{Shift} = d - d'$
$40 = (60 + H) - (50 + \frac{H}{1.6})$
$40 = 10 + H - \frac{H}{1.6}$
$30 = H(1 - \frac{1}{1.6})$
$30 = H(\frac{1.6 - 1}{1.6}) = H(\frac{0.6}{1.6})$
$30 = H(\frac{6}{16}) = H(\frac{3}{8})$
$H = 30 \times \frac{8}{3} = 80 \ cm$.

(B) The apparent depth of the ink mark due to the glass slab is given by $d' = \frac{t}{\mu}$, where $t = 3 \ cm$ is the thickness of the slab and $\mu$ is its refractive index.
The observer is at a distance of $5 \ cm$ above the slab.
Therefore, the total apparent distance of the mark from the observer is the sum of the distance above the slab and the apparent depth of the mark inside the slab.
Total apparent distance = $5 \ cm + \frac{3 \ cm}{\mu} = 4 \ cm$ is incorrect based on the problem statement. Let's re-evaluate: The mark is at the bottom of the slab. The apparent depth of the mark as seen from the top surface is $d_{app} = \frac{t}{\mu} = \frac{3}{\mu}$.
The observer is $5 \ cm$ above the top surface.
So, the total apparent distance from the observer is $5 + \frac{3}{\mu} = 4$ is not possible as the mark must appear closer. The problem implies the total distance from the observer to the apparent position is $4 \ cm$.
If the observer is $5 \ cm$ above the slab, and the mark appears at a distance of $4 \ cm$ from the observer, then the apparent depth $d_{app}$ from the top surface must be $4 - 5 = -1 \ cm$, which is physically impossible.
Re-reading: "For a person looking at the mark at a distance $5.0 \ cm$ above it". This likely means the total distance from the observer to the mark is $5 \ cm$ (including the slab).
Let the total real distance be $5 \ cm$. The slab thickness is $3 \ cm$. The distance above the slab is $5 - 3 = 2 \ cm$.
The apparent depth of the mark from the top surface is $\frac{3}{\mu}$.
The total apparent distance from the observer is $2 + \frac{3}{\mu} = 4 \ cm$.
$\frac{3}{\mu} = 4 - 2 = 2 \ cm$.
$\mu = \frac{3}{2} = 1.5$.

(A) Let the velocity of light in the rarer medium be $v_1$ and in the glass slab be $v_2$. Given that the velocity is reduced by $20 \%$,we have $v_2 = v_1 - 0.20v_1 = 0.8v_1 = \frac{4}{5}v_1$.
From Snell's Law,$n_1 \sin i = n_2 \sin r$. Since $n = \frac{c}{v}$,we have $\frac{c}{v_1} \sin i = \frac{c}{v_2} \sin r$.
This simplifies to $\frac{\sin i}{v_1} = \frac{\sin r}{v_2}$,so $\sin r = \frac{v_2}{v_1} \sin i$.
Given $v_2 = 0.8v_1$,we get $\sin r = 0.8 \sin i$.
For small angles,$\sin \theta \approx \theta$,so $r = 0.8i = \frac{4}{5}i$.
The angle of deviation $\delta$ is given by $\delta = i - r$.
Substituting the value of $r$,we get $\delta = i - 0.8i = 0.2i = \frac{1}{5}i$.

(A) Let the side of the glass cube be $L = 21 \ cm$. Let the actual distance of the air bubble from the first face be $x$. Then,its distance from the opposite face is $(L - x)$.
When viewed from the first face,the apparent distance is $d_1 = x / \mu = 8 \ cm$,so $x = 8\mu$.
When viewed from the opposite face,the apparent distance is $d_2 = (L - x) / \mu = 6 \ cm$,so $L - x = 6\mu$.
Adding the two equations: $x + (L - x) = 8\mu + 6\mu$,which gives $L = 14\mu$.
Given $L = 21 \ cm$,we have $21 = 14\mu$,so $\mu = 21 / 14 = 1.5$.
Substituting $\mu = 1.5$ into $x = 8\mu$,we get $x = 8 \times 1.5 = 12 \ cm$.

(D) The apparent depth of a combination of immiscible liquids is given by the formula: $d_{app} = \sum \frac{d_i}{\mu_i}$,where $d_i$ is the real depth and $\mu_i$ is the refractive index of the $i$-th liquid.
Given:
$d_1 = 3 \text{ cm}, \mu_1 = 3/2$
$d_2 = 4 \text{ cm}, \mu_2 = 4/3$
$d_3 = 6 \text{ cm}, \mu_3 = 6/5$
Calculating the apparent depth:
$d_{app} = \frac{3}{3/2} + \frac{4}{4/3} + \frac{6}{6/5}$
$d_{app} = (3 \times 2/3) + (4 \times 3/4) + (6 \times 5/6)$
$d_{app} = 2 + 3 + 5 = 10 \text{ cm}$.
Therefore,the apparent depth of the vessel is $10 \text{ cm}$.

(B) Let $s_1$ be the distance of the source from the surface and $c$ be the speed of light in air. The time taken to reach the surface is $T_1 = \frac{s_1}{c}$.
Let $s_2$ be the thickness of the glass slab $(5 \ cm)$ and $v$ be the speed of light in glass. The time taken to travel through the slab is $T_2 = \frac{s_2}{v}$.
Given that $T_1 = T_2$,we have $\frac{s_1}{c} = \frac{s_2}{v}$.
Since the refractive index $\mu = \frac{c}{v}$,we can write $v = \frac{c}{\mu}$.
Substituting this into the equation: $\frac{s_1}{c} = \frac{s_2}{c/\mu} = \frac{s_2 \times \mu}{c}$.
Therefore,$s_1 = s_2 \times \mu$.
Given $s_2 = 5 \ cm$ and $\mu = 1.6$,we get $s_1 = 5 \times 1.6 = 8 \ cm$.

(C) Let the height of the water filled in the container be $x$.
The apparent depth of the bottom of the container as seen from the top is given by $d' = \frac{x}{n}$, where $n$ is the refractive index of water.
Given $n = \frac{4}{3}$, the apparent depth is $d' = \frac{x}{4/3} = \frac{3x}{4}$.
The container appears half-filled, which means the apparent depth of the bottom from the top surface of the water is equal to the height of the empty part of the container.
The height of the empty part is $30 - x$.
Therefore, we set $d' = 30 - x$.
$\frac{3x}{4} = 30 - x$
$3x = 120 - 4x$
$7x = 120$
$x = \frac{120}{7} \approx 17.14 \, cm$.
Wait, re-evaluating the condition "appears half-filled": If the total height is $30 \, cm$, half-filled means the apparent position of the bottom is at $15 \, cm$ from the top.
So, $(30 - x) + d' = 15$ is not correct. The condition "appears half-filled" means the apparent depth of the bottom is $15 \, cm$.
$d' = 15 \, cm$.
Since $d' = \frac{x}{n}$, we have $15 = \frac{x}{4/3}$.
$x = 15 \times \frac{4}{3} = 20 \, cm$.
Thus, the height of the water should be $20 \, cm$.

(A) The formula for apparent depth of a single layer is given by $\text{Apparent depth} = \frac{\text{Real depth} (R)}{\text{Refractive index} (\mu)}$.
For a combination of multiple immiscible liquids,the total apparent depth is the sum of the apparent depths of individual layers:
$\text{Total Apparent depth} = \frac{R_1}{\mu_1} + \frac{R_2}{\mu_2} + \frac{R_3}{\mu_3}$
Given:
$R_1 = 3 \ cm, \mu_1 = \frac{3}{2}$
$R_2 = 4 \ cm, \mu_2 = \frac{4}{3}$
$R_3 = 6 \ cm, \mu_3 = \frac{6}{5}$
Substituting these values:
$\text{Apparent depth} = \frac{3}{3/2} + \frac{4}{4/3} + \frac{6}{6/5}$
$= (3 \times \frac{2}{3}) + (4 \times \frac{3}{4}) + (6 \times \frac{5}{6})$
$= 2 + 3 + 5$
$= 10 \ cm$.

(C) Given: The number of waves in glass $(N_g)$ is equal to the number of waves in water $(N_w)$.
$N_g = N_w$
Since the number of waves $N = \frac{d}{\lambda}$,where $d$ is the thickness of the medium and $\lambda$ is the wavelength in that medium:
$\frac{d_g}{\lambda_g} = \frac{d_w}{\lambda_w}$
We know that $\lambda = \frac{\lambda_{\text{air}}}{\mu}$,where $\mu$ is the refractive index.
Substituting this into the equation:
$\frac{d_g \cdot \mu_g}{\lambda_{\text{air}}} = \frac{d_w \cdot \mu_w}{\lambda_{\text{air}}}$
$\mu_g \cdot d_g = \mu_w \cdot d_w$
Given $d_g = 5 \ cm$,$d_w = 6 \ cm$,and $\mu_g = 1.56$:
$1.56 \times 5 = \mu_w \times 6$
$\mu_w = \frac{1.56 \times 5}{6} = \frac{7.8}{6} = 1.30$
Thus,the refractive index of water is $1.30$.

(B) The normal shift $x$ produced by a glass slab of thickness $t$ and refractive index $\mu$ is given by the formula:
$x = t \left(1 - \frac{1}{\mu}\right)$
Rearranging the formula to solve for $t$:
$x = t \left(\frac{\mu - 1}{\mu}\right)$
$t = \frac{x \cdot \mu}{\mu - 1}$
Therefore,the thickness of the glass slab is $t = \frac{\mu x}{\mu - 1}$.

(C) The apparent depth of a vessel containing multiple layers of non-mixing liquids is given by the sum of the apparent depths of each individual layer.
The formula for apparent depth is $d_{app} = \sum \frac{d_i}{n_i}$,where $d_i$ is the real depth and $n_i$ is the refractive index of the $i$-th layer.
Given:
Layer $1$: $d_1 = 40 \ cm$,$n_1 = 1.6$
Layer $2$: $d_2 = 30 \ cm$,$n_2 = 1.5$
Calculating the apparent depth:
$d_{app} = \frac{40}{1.6} + \frac{30}{1.5}$
$d_{app} = 25 \ cm + 20 \ cm$
$d_{app} = 45 \ cm$
Therefore,the apparent depth of the vessel is $45 \ cm$.

(A) The formula for the apparent shift $(h)$ due to a glass slab is given by:
$h = t \left(1 - \frac{1}{n}\right)$
where $t$ is the real thickness of the slab and $n$ is the refractive index.
Given:
$t = 4.8 \text{ cm}$
$n = 1.5$
Substituting the values:
$h = 4.8 \left(1 - \frac{1}{1.5}\right)$
$h = 4.8 \left(1 - \frac{2}{3}\right)$
$h = 4.8 \left(\frac{1}{3}\right)$
$h = 1.6 \text{ cm}$
Therefore, the ink dot appears to be raised by $1.6 \text{ cm}$.

(B) Let the total depth of the liquid in the tank be $H$. The object $P$ is at a height $h$ above the mirror,so its depth from the liquid surface is $(H-h)$.
The apparent depth of the object $P$ as seen by the observer $O$ is $d_1 = \frac{H-h}{\mu}$.
The image of the object $P$ formed by the plane mirror is at a distance $h$ below the mirror. The total depth of this image from the liquid surface is $(H+h)$.
The apparent depth of the image as seen by the observer $O$ is $d_2 = \frac{H+h}{\mu}$.
The apparent distance between the object $P$ and its image is the difference between their apparent depths:
$\text{Apparent distance} = d_2 - d_1 = \frac{H+h}{\mu} - \frac{H-h}{\mu} = \frac{H+h-H+h}{\mu} = \frac{2h}{\mu}$.

(B) For small angles,Snell's law is given by $n = \frac{\sin i}{\sin r} \approx \frac{i}{r}$.
Given that the velocity is reduced by $25 \%$,the new velocity $v = c - 0.25c = 0.75c = \frac{3}{4}c$.
The refractive index $n$ is defined as $n = \frac{c}{v} = \frac{c}{0.75c} = \frac{1}{0.75} = \frac{4}{3}$.
Equating the two expressions for $n$: $\frac{i}{r} = \frac{4}{3} \implies r = \frac{3}{4}i$.
The angle of deviation $\delta$ is given by $\delta = i - r$.
Substituting the value of $r$: $\delta = i - \frac{3}{4}i = \frac{i}{4}$.

(A) Let the distance of the air bubble from the first surface be $l_1$ and from the second surface be $l_2$. The total length of the cube is $L = 24 \,cm$. Thus, $l_1 + l_2 = 24 \,cm$, which implies $l_2 = 24 - l_1$.
The formula for apparent depth is given by $\mu = \frac{\text{Real depth}}{\text{Apparent depth}}$.
For the first surface:
$\mu = \frac{l_1}{10} \quad \dots(i)$
For the second surface:
$\mu = \frac{l_2}{6} = \frac{24 - l_1}{6} \quad \dots(ii)$
Equating $(i)$ and $(ii)$:
$\frac{l_1}{10} = \frac{24 - l_1}{6}$
$6l_1 = 10(24 - l_1)$
$6l_1 = 240 - 10l_1$
$16l_1 = 240$
$l_1 = \frac{240}{16} = 15 \,cm$
Therefore, the distance of the air bubble from the first surface is $15 \,cm$.

(C) The correct option is $(C)$.
Let the real depth of the fish be $h$ and the apparent depth be $h'$.
From the geometry of refraction at a plane surface, for small angles of incidence $i$ and refraction $r$, we have:
$\sin(i) \approx \tan(i) = \frac{P}{h}$
$\sin(r) \approx \tan(r) = \frac{P}{h'}$
According to Snell's Law: $n_1 \sin(i) = n_2 \sin(r)$
Here, $n_1 = \mu = \frac{4}{3}$ (water) and $n_2 = 1$ (air).
So, $\mu \times \frac{P}{h} = 1 \times \frac{P}{h'}$
This simplifies to $h = \mu h'$.
Given $h' = 30 \,cm$ and $\mu = \frac{4}{3}$, we get:
$h = \frac{4}{3} \times 30 \,cm = 40 \,cm$.
Thus, the real depth of the fish is $40 \,cm$.

(A) The number of waves $N$ in a medium of thickness $t$ is given by $N = \frac{t}{\lambda_m}$, where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium and $\lambda_0$ is the wavelength in vacuum.
Thus, $N = \frac{t \cdot \mu}{\lambda_0}$.
Given that the number of waves in the glass slab $(t_g = 4 \,cm, \mu_g = 5/3)$ is equal to the number of waves in the water column $(t_w = 5 \,cm, \mu_w = ?)$, we have:
$\frac{t_g \cdot \mu_g}{\lambda_0} = \frac{t_w \cdot \mu_w}{\lambda_0}$
$t_g \cdot \mu_g = t_w \cdot \mu_w$
Substituting the given values:
$4 \times \frac{5}{3} = 5 \times \mu_w$
$\frac{20}{3} = 5 \times \mu_w$
$\mu_w = \frac{20}{3 \times 5} = \frac{4}{3}$.

(C) Let the thickness of the glass slab be $t$ and the actual distance of the bubble from one side be $x$.
When viewed from one side, the apparent depth is given by $d_1 = x / \mu = 5 \,cm$.
When viewed from the other side, the apparent depth is given by $d_2 = (t - x) / \mu = 2 \,cm$.
Adding these two equations:
$d_1 + d_2 = (x / \mu) + ((t - x) / \mu) = t / \mu$.
Given $\mu = 1.5$, $d_1 = 5 \,cm$, and $d_2 = 2 \,cm$:
$5 + 2 = t / 1.5$.
$7 = t / 1.5$.
$t = 7 \times 1.5 = 10.5 \,cm$.
Therefore, the thickness of the slab is $10.5 \,cm$.

(A) Let the number of waves be $N$. The wavelength in a medium is given by $\lambda = \frac{d}{N}$,where $d$ is the thickness of the medium.
For the glass slab,$\lambda_g = \frac{4}{N}$.
For the water column,$\lambda_w = \frac{5}{N}$.
We know that the wavelength in a medium is inversely proportional to its refractive index,i.e.,$\lambda \propto \frac{1}{\mu}$.
Therefore,$\frac{\lambda_w}{\lambda_g} = \frac{\mu_g}{\mu_w}$.
Substituting the values: $\frac{5/N}{4/N} = \frac{5/3}{\mu_w}$.
$\frac{5}{4} = \frac{5}{3 \mu_w}$.
$3 \mu_w = 4$.
$\mu_w = \frac{4}{3} \approx 1.33$.

(B) Let the distance of the source from the surface be $x$. The thickness of the glass slab is $d = 5 \ cm$,and its refractive index is $\mu = 1.6$.
The time taken by light to travel a distance $x$ in air (where speed is $c$) is $t_1 = \frac{x}{c}$.
The time taken by light to travel through the glass slab of thickness $d$ (where speed is $v = \frac{c}{\mu}$) is $t_2 = \frac{d}{v} = \frac{d}{c/\mu} = \frac{\mu d}{c}$.
According to the problem,$t_1 = t_2$,so:
$\frac{x}{c} = \frac{\mu d}{c}$
$x = \mu d$
$x = 1.6 \times 5 \ cm = 8 \ cm$.
Therefore,the distance of the source from the surface is $8 \ cm$.

(C) The number of waves $N$ in a medium of thickness $t$ is given by $N = \frac{t}{\lambda_m}$,where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium and $\lambda_0$ is the wavelength in vacuum.
Thus,$N = \frac{t \cdot \mu}{\lambda_0}$.
Given that the number of waves in the glass slab is equal to the number of waves in the water column:
$\frac{t_g \cdot \mu_g}{\lambda_0} = \frac{t_w \cdot \mu_w}{\lambda_0}$
$\therefore \mu_g \cdot t_g = \mu_w \cdot t_w$
Substituting the given values: $\mu_g = 1.5$,$t_g = 6 \ cm$,and $t_w = 7 \ cm$:
$1.5 \times 6 = \mu_w \times 7$
$9 = 7 \cdot \mu_w$
$\mu_w = \frac{9}{7} \approx 1.286$.

(B) The refractive index $n$ is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium,$n = \frac{c}{v}$.
Given that the velocity is reduced by $20 \%$,the new velocity $v = c - 0.20c = 0.80c = \frac{4}{5}c$.
Thus,the refractive index $n = \frac{c}{v} = \frac{c}{0.8c} = \frac{1}{0.8} = 1.25$ or $\frac{5}{4}$.
According to Snell's Law for small angles,$n_1 \sin i = n_2 \sin r$. Since the incident medium is air $(n_1 = 1)$,we have $i = n r$,so $r = \frac{i}{n} = \frac{i}{1.25} = 0.8i = \frac{4i}{5}$.
The angle of deviation $\delta$ is given by $\delta = i - r$.
Substituting the value of $r$,we get $\delta = i - 0.8i = 0.2i = \frac{i}{5}$.

(A) The number of waves $N$ in a medium of thickness $t$ is given by $N = t / \lambda_m$,where $\lambda_m = \lambda_0 / n$ is the wavelength in the medium and $\lambda_0$ is the wavelength in vacuum.
Thus,$N = (t \cdot n) / \lambda_0$.
Since the number of waves is the same for both glass and water:
$N_{glass} = N_{water}$
$(t_g \cdot n_g) / \lambda_0 = (t_w \cdot n_w) / \lambda_0$
$t_g \cdot n_g = t_w \cdot n_w$
Given $t_g = 4 \ cm$,$t_w = 5 \ cm$,and $n_w = 4/3$:
$4 \cdot n_g = 5 \cdot (4/3)$
$n_g = (5 \cdot 4) / (3 \cdot 4)$
$n_g = 5/3$.

(D) The speed of light in a medium with refractive index $\mu$ is given by $v = \frac{c}{\mu}$,where $c$ is the speed of light in a vacuum.
Time taken $(T)$ to travel a distance $t$ is given by the formula $T = \frac{\text{distance}}{\text{speed}}$.
Substituting the values,we get $T = \frac{t}{v} = \frac{t}{(c/\mu)}$.
Therefore,$T = \frac{\mu t}{c}$.

(C) The number of waves $N$ in a medium of thickness $d$ and refractive index $\mu$ is given by $N = \frac{d}{\lambda_m}$,where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium.
Thus,$N = \frac{d \cdot \mu}{\lambda_0}$.
Since the number of waves is the same for both media,we have:
$\frac{d_g \cdot \mu_g}{\lambda_0} = \frac{d_w \cdot \mu_w}{\lambda_0}$
$d_g \cdot \mu_g = d_w \cdot \mu_w$
Given $d_g = 4 \,cm$,$\mu_g = \frac{5}{3}$,and $\mu_w = \frac{4}{3}$.
Substituting the values: $4 \times \frac{5}{3} = x \times \frac{4}{3}$
$\frac{20}{3} = \frac{4x}{3}$
$4x = 20$
$x = 5 \,cm$.

(B) The refractive index $n$ is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$,given by $n = \frac{c}{v}$.
Since the speed $v$ is the distance $d$ traveled per unit time $t$,we have $v = \frac{d}{t}$.
Substituting this into the refractive index formula: $n = \frac{c}{d/t} = \frac{ct}{d}$.
Rearranging to solve for time $t$: $t = \frac{nd}{c}$.
Given: $n = 1.5$,$d = 4 \ cm = 4 \times 10^{-2} \ m$,and $c = 3 \times 10^{8} \ m/s$.
Substituting the values: $t = \frac{1.5 \times 4 \times 10^{-2}}{3 \times 10^{8}}$.
$t = \frac{6 \times 10^{-2}}{3 \times 10^{8}} = 2 \times 10^{-10} \ s$.

(B) The apparent depth of an object viewed through a glass slab is given by the formula:
$\text{Apparent depth} = \frac{\text{Real depth}}{\mu}$
Given, real depth $(t)$ = $9 \ cm$ and refractive index $(\mu)$ = $1.5$.
$\text{Apparent depth} = \frac{9}{1.5} = 6 \ cm$
The shift or the distance by which the pin appears to be raised is given by:
$\text{Shift} = \text{Real depth} - \text{Apparent depth}$
$\text{Shift} = 9 \ cm - 6 \ cm = 3 \ cm$
Therefore, the pin appears to be raised by $3 \ cm$.

(A) The refractive index $n$ is given by the ratio of real depth to apparent depth.
Real depth $(RD)$ is the actual thickness of the glass slab,which is the difference between the reading for the chalk dust on top of the slab and the reading for the ink mark at the bottom: $RD = 8.123 \ cm - 5.123 \ cm = 3.000 \ cm$.
Apparent depth $(AD)$ is the depth of the ink mark as seen through the glass slab,which is the difference between the reading for the chalk dust on top and the reading for the ink mark through the slab: $AD = 8.123 \ cm - 6.123 \ cm = 2.000 \ cm$.
The refractive index is $n = \frac{RD}{AD} = \frac{3.000}{2.000} = 1.5$.

(D) The refractive index of a glass plate varies for different colours because the wavelength of light is different for each colour.
According to Cauchy's equation,the refractive index is higher for shorter wavelengths.
Since violet light has the shortest wavelength,the refractive index of glass is maximum for violet colour and minimum for red colour.
The apparent shift in the position of an object viewed through a glass slab is given by the formula: $\Delta t = t(1 - \frac{1}{\mu})$,where $t$ is the thickness of the slab and $\mu$ is the refractive index.
Alternatively,using apparent depth: $d' = \frac{d}{\mu}$. The shift is $\Delta d = d - d' = d(1 - \frac{1}{\mu})$.
Since $\mu$ is maximum for violet light,the term $(1 - \frac{1}{\mu})$ is maximum for violet.
Therefore,the violet letter appears to be raised the most.

(C) The lateral shift is given by $L_{s} = t \frac{\sin(i-r)}{\cos r}$. As the angle of incidence $i$ increases,the lateral shift $L_{s}$ increases. Thus,statement $A$ is correct.
As the refractive index $\mu$ increases,the angle of refraction $r$ decreases,which leads to an increase in the lateral shift $L_{s}$. Thus,statement $B$ is correct.
The normal shift is given by $L_{N} = t(1 - \frac{1}{\mu})$. As the refractive index $\mu$ increases,the term $\frac{1}{\mu}$ decreases,which means $(1 - \frac{1}{\mu})$ increases. Therefore,the normal shift $L_{N}$ increases as the refractive index increases. Thus,statement $C$ is incorrect.
Both $L_{s}$ and $L_{N}$ are directly proportional to the thickness $t$ of the medium. Thus,statement $D$ is correct.
Therefore,the wrong statement is $C$.

(B) The speed of light in a medium is given by $v = c/n$,where $c$ is the speed of light in vacuum and $n$ is the refractive index of the medium.
According to Cauchy's formula,the refractive index $n$ of a material depends on the wavelength $\lambda$ of light.
For glass,the refractive index is highest for violet light and lowest for red light.
Since $v \propto 1/n$,the speed of light is inversely proportional to the refractive index.
Therefore,because red light has the lowest refractive index in glass,it travels with the maximum speed compared to other colours.

(B) The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum $(c)$ to the speed of light in the medium $(v)$: $n = c/v$.
Given, $n = 1.5$ and $c = 3 \times 10^{8} \,m/s$.
Therefore, the speed of light in the glass slab is $v = c/n = (3 \times 10^{8}) / 1.5 = 2 \times 10^{8} \,m/s$.
The thickness of the glass slab is $d = 4 \,mm = 4 \times 10^{-3} \,m$.
The time $t$ taken to pass through the slab is given by $t = d/v$.
Substituting the values: $t = (4 \times 10^{-3} \,m) / (2 \times 10^{8} \,m/s) = 2 \times 10^{-11} \,s$.

(B) Given: Angle of incidence $i = 45^{\circ}$.
Lateral shift per unit thickness $\frac{d}{t} = \frac{1}{\sqrt{3}}$.
The formula for lateral shift $d$ is given by $d = \frac{t \sin(i - r)}{\cos r}$,where $t$ is the thickness and $r$ is the angle of refraction.
Rearranging the formula: $\frac{d}{t} = \frac{\sin(i - r)}{\cos r}$.
Expanding $\sin(i - r)$: $\frac{d}{t} = \frac{\sin i \cos r - \cos i \sin r}{\cos r} = \sin i - \cos i \tan r$.
Substituting the values: $\frac{1}{\sqrt{3}} = \sin 45^{\circ} - \cos 45^{\circ} \tan r$.
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have: $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{2}}(1 - \tan r)$.
Multiplying by $\sqrt{2}$: $\frac{\sqrt{2}}{\sqrt{3}} = 1 - \tan r$.
Therefore,$\tan r = 1 - \sqrt{\frac{2}{3}}$.
Thus,$r = \tan^{-1}\left(1 - \sqrt{\frac{2}{3}}\right)$.

(D) The formula for apparent depth when viewed normally from a rarer medium is given by: $\text{Apparent depth} = \frac{\text{Real depth}}{\mu}$.
Given,$\text{Real depth} = 100 \ cm$ and refractive index $\mu = \frac{4}{3}$.
Substituting the values: $\text{Apparent depth} = \frac{100}{4/3} = 100 \times \frac{3}{4} = 75 \ cm$.
Therefore,the apparent depth of the object is $75 \ cm$.

(B) When a vessel is filled with multiple immiscible liquids of refractive indices $\mu_1, \mu_2, \dots, \mu_n$ and real depths $d_1, d_2, \dots, d_n$,the total apparent depth $d_{app}$ when viewed perpendicularly is given by the formula:
$d_{app} = \sum_{i=1}^{n} \frac{d_i}{\mu_i} = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} + \dots + \frac{d_n}{\mu_n}$
In this problem,the total depth is $2d \text{ cm}$,which is divided into two equal halves. Thus,the real depth of each liquid is $d_1 = d$ and $d_2 = d$.
The refractive index of the lower half is $\mu_1$ and the upper half is $\mu_2$.
Substituting these values into the formula:
$d_{app} = \frac{d}{\mu_1} + \frac{d}{\mu_2}$
$d_{app} = d\left(\frac{1}{\mu_1} + \frac{1}{\mu_2}\right)$

(B) We know that the refractive index is given by the formula: $\mu = \frac{\text{Real depth}}{\text{Apparent depth}}$.
Here,$\mu = 1.5$ and the real depth is the thickness of the glass slab,which is $0.12 \,m$.
Therefore,the apparent depth is calculated as: $\text{Apparent depth} = \frac{\text{Real depth}}{\mu} = \frac{0.12}{1.5} = 0.08 \,m$.
The shift in the position of the ink dot is given by: $\text{Shift} = \text{Real depth} - \text{Apparent depth} = 0.12 \,m - 0.08 \,m = 0.04 \,m$.
Since the image of the dot appears to be raised by $0.04 \,m$,the travelling microscope must be moved upwards by $0.04 \,m$ to focus on the dot again.

(B) Let $u_1$ be the initial object distance from the lens and $f$ be the focal length of the lens.
Given: $v_1 = 60 \ cm$ (initial image distance) and $m = 2$ (lateral magnification).
Using the magnification formula $m = \frac{v_1}{u_1} = 2$ (for a real image formed by a convex lens,$v$ is positive and $u$ is negative,so $m = v/u$ in magnitude for the distance ratio),we have:
$\frac{60}{|u_1|} = 2 \Rightarrow |u_1| = 30 \ cm$. Thus,$u_1 = -30 \ cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v_1} - \frac{1}{u_1}$:
$\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1+2}{60} = \frac{3}{60} = \frac{1}{20} \Rightarrow f = 20 \ cm$.
When liquid of height $H = 24 \ cm$ is filled,the object appears to shift upwards due to refraction. The new apparent object distance $u_2$ is related to the original distance by the normal shift formula: $u_2 = |u_1| - H(1 - \frac{1}{\mu})$.
Using the lens formula for the new image distance $v_2 = 120 \ cm$:
$\frac{1}{f} = \frac{1}{v_2} - \frac{1}{u_2} \Rightarrow \frac{1}{20} = \frac{1}{120} - \frac{1}{u_2} \Rightarrow \frac{1}{u_2} = \frac{1}{120} - \frac{1}{20} = \frac{1-6}{120} = -\frac{5}{120} = -\frac{1}{24}$.
So,$|u_2| = 24 \ cm$.
Equating the shift: $|u_1| - |u_2| = H(1 - \frac{1}{\mu})$
$30 - 24 = 24(1 - \frac{1}{\mu}) \Rightarrow 6 = 24(1 - \frac{1}{\mu})$
$1 - \frac{1}{\mu} = \frac{6}{24} = \frac{1}{4} \Rightarrow \frac{1}{\mu} = 1 - \frac{1}{4} = \frac{3}{4} \Rightarrow \mu = \frac{4}{3} \approx 1.33$.

(B) The sun is at infinity, so $u = \infty$.
The light is focused at a height $v = 32 \,cm$ from the mirror. Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{32} + \frac{1}{\infty} \Rightarrow f = 32 \,cm$.
When the tank is filled with water up to a height $h = 20 \,cm$, the light rays are refracted at the water surface.
The mirror would have formed an image at $32 \,cm$ from the bottom. Since the water level is at $20 \,cm$, the distance of this image from the water surface is $BO = 32 \,cm - 20 \,cm = 12 \,cm$.
This image acts as a virtual object for the water surface. Due to refraction, the apparent height $BI$ from the water surface is given by:
$BI = \frac{BO}{\mu} = \frac{12}{4/3} = 12 \times \frac{3}{4} = 9 \,cm$.
Thus, the sunlight focuses at $9 \,cm$ above the water level.

(C) According to Snell's law,$\sin r = \frac{\sin i}{\mu} = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$.
Therefore,the angle of refraction $r = 30^{\circ}$.
The lateral shift $d$ produced by a glass slab of thickness $b$ is given by the formula:
$d = b \cdot \frac{\sin(i - r)}{\cos r}$
Substituting the given values:
$d = 0.04 \cdot \frac{\sin(60^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$
$d = 0.04 \cdot \frac{\sin 30^{\circ}}{\cos 30^{\circ}} = 0.04 \cdot \tan 30^{\circ}$
$d = 0.04 \cdot \frac{1}{\sqrt{3}} \text{ m}$
To convert the result into $\text{mm}$,we multiply by $1000$:
$d = \frac{0.04 \times 1000}{\sqrt{3}} \text{ mm} = \frac{40}{\sqrt{3}} \text{ mm}$.