Critical Angle and Total Internal Reflection Questions in English

(A) For total internal reflection to occur at the interface between the glass prism and water,the angle of incidence $\theta$ must be greater than or equal to the critical angle $C$.
The condition for total internal reflection is $\sin \theta \geqslant \sin C$.
The critical angle $C$ is given by $\sin C = \frac{1}{{}_w\mu_g}$,where ${}_w\mu_g$ is the refractive index of glass with respect to water.
Given,refractive index of glass $\mu_g = 1.5 = 3/2$ and refractive index of water $\mu_w = 4/3$.
Therefore,${}_w\mu_g = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
Thus,$\sin C = \frac{1}{9/8} = \frac{8}{9}$.
For total internal reflection,the angle of incidence $\theta$ must satisfy $\sin \theta \geqslant \frac{8}{9}$.

(D) For total internal reflection to occur at the prism-liquid interface,the angle of incidence $i$ must be greater than or equal to the critical angle $c$.
In the given figure,the ray strikes the interface at an angle of incidence $i = 45^o$.
For total internal reflection,we require $i \ge c$,so $45^o \ge c$,or $\sin 45^o \ge \sin c$.
Since $\sin c = \frac{\mu_{liquid}}{\mu_{prism}}$,and assuming the prism is glass with $\mu_{prism} \approx 1.5$,the condition for total internal reflection is $\sin 45^o \ge \frac{\mu_{liquid}}{\mu_{prism}}$.
However,the question asks for the maximum refractive index of the liquid such that total internal reflection still occurs. This occurs when $c = 45^o$.
Using the relation $\sin c = \frac{\mu_{liquid}}{\mu_{prism}}$,we get $\mu_{liquid} = \mu_{prism} \sin 45^o = 1.5 \times \frac{1}{\sqrt 2} \approx 1.06$.
The assertion states the value is $\sqrt 2$,which is incorrect as it assumes the prism refractive index is $2$. Given the standard context of such problems,the assertion and reason are both technically incorrect based on standard glass prisms.

(C) Given: Refractive index of diamond with respect to air,$\mu_d = \sqrt{6}$. Refractive index of liquid with respect to air,$\mu_l = \sqrt{3}$.
The refractive index of diamond with respect to the liquid is given by $\mu = \frac{\mu_d}{\mu_l} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}$.
For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $C$. The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$.
Thus,$C = 45^{\circ}$.
Since the given angle of incidence $i = 30^{\circ}$ is less than the critical angle $C = 45^{\circ}$,total internal reflection will not occur. Therefore,the Assertion is incorrect.
The Reason states $\mu = \frac{1}{\sin C}$,which is the correct formula for the critical angle where $\mu$ is the refractive index of the denser medium with respect to the rarer medium. Thus,the Reason is correct.

(C) An air bubble in water acts as a spherical lens-like boundary between a denser medium (water) and a rarer medium (air).
When light rays travel from water to the air bubble,they strike the interface at an angle greater than the critical angle for the water-air interface.
This leads to the phenomenon of total internal reflection,which causes the bubble to appear shiny or silvery.
Refraction is not the primary cause of this phenomenon; rather,it is total internal reflection.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) The Assertion is correct because diamond has a very high refractive index $(n \approx 2.42)$,which results in a very small critical angle $(C \approx 24.4^\circ)$.
When light enters a diamond,it undergoes multiple internal reflections due to this small critical angle,causing it to glitter brilliantly.
The Reason is incorrect because the glittering of a diamond is not related to the absorption of sunlight,but rather to the phenomenon of Total Internal Reflection $(TIR)$.

(A) The Assertion and Reason are both correct,and the Reason provides a correct explanation for the Assertion.
In an optical fibre,the core diameter is kept small to ensure that the light rays entering the fibre strike the core-cladding interface at an angle greater than the critical angle.
As shown in the diagram,for a small diameter core,the angle of incidence $\angle A$ at the core-cladding interface is relatively large.
For a larger diameter core,the angle of incidence $\angle B$ is smaller.
Since the condition for total internal reflection is that the angle of incidence must be greater than the critical angle,a smaller core diameter increases the probability that the light rays will satisfy this condition,thereby ensuring efficient transmission of light through the fibre.

(A) The critical angle $\theta_c$ is given by the formula $\theta_c = \sin^{-1} \left( \frac{1}{\mu} \right)$.
According to Cauchy's dispersion formula,the refractive index $\mu$ of a medium is inversely proportional to the square of the wavelength $\lambda$ (approximately $\mu \propto \frac{1}{\lambda}$ for visible light).
Since the wavelength of violet light is the minimum among visible colors,the refractive index $\mu$ for violet light is the maximum.
As $\theta_c = \sin^{-1} \left( \frac{1}{\mu} \right)$,a higher value of $\mu$ results in a smaller value of $\theta_c$.
Therefore,the critical angle is minimum for violet color. Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(A) The Assertion is correct because optical fibres are widely used in modern telecommunication systems for high-speed data transmission.
The Reason is also correct because the fundamental principle behind the operation of an optical fibre is the phenomenon of total internal reflection $(TIR)$.
When light enters the fibre,it undergoes multiple total internal reflections at the core-cladding interface,allowing the signal to travel over long distances with minimal loss of intensity.
Since the ability to transmit signals over long distances without significant attenuation is directly due to $TIR$,the Reason is the correct explanation of the Assertion.

(D) The critical angle $(i_c)$ is defined as the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90^o$.
When the light ray travels from a denser medium to a rarer medium at an angle of incidence equal to the critical angle,the refracted ray grazes the interface between the two media.
Therefore,the angle of refraction is $90^o$.

(D) The refractive index $\mu$ of a medium is given by $\mu = \sqrt{\epsilon_r \mu_r}$.
Given relative permittivity $\epsilon_r = 3$ and relative permeability $\mu_r = \frac{4}{3}$.
Therefore,$\mu = \sqrt{3 \times \frac{4}{3}} = \sqrt{4} = 2$.
The critical angle $\theta_C$ is given by $\sin \theta_C = \frac{1}{\mu}$.
Substituting the value of $\mu$,we get $\sin \theta_C = \frac{1}{2}$.
Thus,$\theta_C = 30^{\circ}$.

(A) The light source emits light in all directions,covering a total solid angle of $4\pi$ steradians.
Light emerges from the water surface only if the angle of incidence is less than or equal to the critical angle $\theta_c$.
Using Snell's law at the critical angle: $\mu \sin\theta_c = 1 \sin 90^{\circ}$,where $\mu = 4/3$.
$\sin\theta_c = 1 / (4/3) = 3/4$.
Then,$\cos\theta_c = \sqrt{1 - \sin^2\theta_c} = \sqrt{1 - 9/16} = \sqrt{7}/4$.
The solid angle $\Omega$ subtended by the cone of light that emerges is $\Omega = 2\pi(1 - \cos\theta_c)$.
Substituting the value of $\cos\theta_c$: $\Omega = 2\pi(1 - \sqrt{7}/4) = 2\pi(1 - 2.646/4) = 2\pi(1 - 0.6615) = 2\pi(0.3385) = 0.677\pi$.
The fraction of light emerging is $\frac{\Omega}{4\pi} = \frac{0.677\pi}{4\pi} \approx 0.169$.
Converting to percentage: $0.169 \times 100 \approx 17\%$.

(B) The depth of the bulb in water is $h = 80 \; cm = 0.8 \; m$.
The refractive index of water is $\mu = 1.33$.
Light emerges from the surface only if the angle of incidence is less than or equal to the critical angle $i_c$. At the critical angle,the angle of refraction is $r = 90^{\circ}$.
Using Snell's law: $\mu \sin i_c = 1 \sin 90^{\circ} \implies \sin i_c = \frac{1}{\mu} = \frac{1}{1.33} \approx 0.7519$.
From the geometry,the radius $R$ of the circular area on the surface is given by $R = h \tan i_c$.
Since $\sin i_c = 0.7519$,we have $\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - (0.7519)^2} \approx 0.6593$.
Thus,$\tan i_c = \frac{\sin i_c}{\cos i_c} = \frac{0.7519}{0.6593} \approx 1.1404$.
The radius $R = 0.8 \times 1.1404 \approx 0.9123 \; m$.
The area of the circular surface is $A = \pi R^2 = \pi \times (0.9123)^2 \approx 3.14159 \times 0.8323 \approx 2.61 \; m^2$.

(N/A) Refractive index of the glass fibre,$\mu_{1} = 1.68$.
Refractive index of the outer covering of the pipe,$\mu_{2} = 1.44$.
Let $i$ be the angle of incidence at the air-core interface and $r$ be the angle of refraction.
Let $i'$ be the angle of incidence at the core-cladding interface.
For total internal reflection $(TIR)$ to occur at the core-cladding interface,the angle of incidence $i'$ must be greater than the critical angle $i_c$.
$\sin i_c = \frac{\mu_2}{\mu_1} = \frac{1.44}{1.68} \approx 0.8571$.
Thus,$i_c = \sin^{-1}(0.8571) \approx 59^{\circ}$.
For $TIR$,$i' > 59^{\circ}$.
In the right-angled triangle formed at the entry,$r = 90^{\circ} - i'$.
Since $i' > 59^{\circ}$,$r < 90^{\circ} - 59^{\circ} = 31^{\circ}$.
Applying Snell's law at the air-core interface: $\frac{\sin i}{\sin r} = \mu_1 = 1.68$.
$\sin i = 1.68 \sin r$.
For maximum $i$,$r = 31^{\circ}$.
$\sin i_{\max} = 1.68 \sin 31^{\circ} = 1.68 \times 0.515 = 0.8652$.
$i_{\max} = \sin^{-1}(0.8652) \approx 60^{\circ}$.
So,the range of incident angles is $0 < i < 60^{\circ}$.
$(b)$ If there is no outer covering,the cladding is air,so $\mu_2 = 1.0$.
$\sin i_c = \frac{1.0}{1.68} \approx 0.5952$.
$i_c = \sin^{-1}(0.5952) \approx 36.5^{\circ}$.
For $TIR$,$i' > 36.5^{\circ}$.
$r < 90^{\circ} - 36.5^{\circ} = 53.5^{\circ}$.
$\sin i_{\max} = 1.68 \sin 53.5^{\circ} = 1.68 \times 0.8038 \approx 1.35$.
Since $\sin i$ cannot exceed $1$,$i_{\max} = 90^{\circ}$.
Thus,all rays entering the pipe will undergo $TIR$.

(N/A) When light travels from an optically denser medium to a rarer medium at the interface,it is partly reflected back into the same medium and partly refracted into the second medium. This reflection is called internal reflection.
When a ray of light enters from a denser medium to a rarer medium,it bends away from the normal.
In the figure,for the incident ray $AO_{1}$,the light is partially reflected $(O_{1}C)$ and partially transmitted $(O_{1}B)$ or refracted.
The angle of refraction $(r)$ is larger than the angle of incidence $(i)$.
As the angle of incidence increases,so does the angle of refraction,until for the ray $AO_{3}$,the angle of refraction becomes $\frac{\pi}{2}$.
The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. This is shown by the ray $AO_{3}D$.
The angle of incidence for which the angle of refraction becomes $\frac{\pi}{2}$ is called the critical angle '$i_{C}$'.
The deviated ray obtained after incidence at the critical angle is called the critical ray.
If the angle of incidence is increased further,e.g.,for the ray $AO_{4}$,refraction is not possible and the incident ray is totally reflected. This is called total internal reflection.
When light gets reflected by a surface,normally some fraction of it gets transmitted. The reflected ray,therefore,is always less intense than the incident ray,no matter how smooth the reflecting surface may be. In total internal reflection,on the other hand,no transmission of light takes place.

(N/A) $1$. Take a glass beaker filled with clear water. Stir the water with a small piece of soap to make it slightly turbid so that the path of the light beam becomes visible.
$2$. Use a laser pointer to shine a beam through the turbid water. The path of the beam inside the water will shine brightly.
$3$. Shine the beam from below the beaker such that it strikes the upper water surface. At this point,it undergoes partial reflection (seen as a spot on the table below) and partial refraction (seen as a spot on the ceiling),as shown in figure $(a)$.
$4$. Now,direct the laser beam from one side of the beaker so that it strikes the upper surface of the water more obliquely,as shown in figure $(b)$. Adjust the angle of the laser beam until the refraction above the water surface disappears entirely,and the beam is completely reflected back into the water. This phenomenon is total internal reflection.
$5$. Pour this water into a long test tube and shine the laser light from the top,as shown in figure $(c)$. Adjust the direction of the laser beam so that it undergoes total internal reflection every time it strikes the walls of the tube. This demonstrates the principle used in optical fibres.
Note: Do not look directly into the laser beam and avoid pointing it at anyone's face.

(D) Total internal reflection $(TIR)$ occurs when light travels from a denser medium to a rarer medium at an angle of incidence greater than the critical angle.
$1$. Mirage: It is an optical illusion caused by the refraction of light in layers of air with varying temperatures,leading to $TIR$.
$2$. Prism: Prisms can be designed to act as total reflecting prisms using $TIR$ to deviate light by $90^{\circ}$ or $180^{\circ}$.
$3$. Diamond: The brilliance of a diamond is due to multiple $TIR$ of light inside it,as its critical angle is very small.
$4$. Optical fibres: These use $TIR$ to transmit light signals over long distances without significant loss of intensity.
Therefore,all the given phenomena are based on $TIR$.

On hot summer days,the air near the ground becomes hotter than the air at higher levels.
The refractive index of air increases with its density.
Hotter air is less dense and has a smaller refractive index than cooler air.
If the air currents are small,that is,the air is still,the optical density at different layers of air increases with height.
As a result,light from a tall object,such as a tree,passes through a medium whose refractive index decreases towards the ground. Thus,a ray of light from such an object successively bends away from the normal and undergoes total internal reflection if the angle of incidence for the air near the ground exceeds the critical angle.
To a distant observer,the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground,perhaps by a pool of water near the tall object. Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called a mirage.
This type of mirage is especially common in hot deserts.
While moving in a bus or a car during a hot summer day,a distant patch of road,especially on a highway,appears to be wet. But you do not find any evidence of wetness when you reach that spot. This is also due to a mirage.

(B) Diamonds are known for their spectacular brilliance. Their brilliance is mainly due to the total internal reflection of light inside them.
The critical angle for a diamond-air interface $\left(\cong 24.4^{\circ}\right)$ is very small. Therefore,once light enters a diamond,it is very likely to undergo total internal reflection inside it.
$\therefore$ For total internal reflection,
$\sin i_{c} = \frac{1}{n}$
$\therefore n = \frac{1}{\sin 24.4^{\circ}}$
$\therefore n = \frac{1}{0.4131}$
$\therefore n \approx 2.42$
Because the refractive index of diamond is very high,the critical angle is very small,which allows light to be trapped and reflected multiple times,resulting in its characteristic brilliance.

(N/A) Prisms designed to deviate light by $90^{\circ}$ or $180^{\circ}$ utilize the phenomenon of total internal reflection,as shown in figures $(a)$ and $(b)$.
In these cases,the angle of incidence at the prism face is $45^{\circ}$. For total internal reflection to occur,the critical angle $i_{c}$ for the material of the prism must be less than $45^{\circ}$.
Such prisms are also used to invert images without changing their size,as illustrated in figure $(c)$.

(N/A) Principle: Optical fibres are based on the phenomenon of total internal reflection.
Construction: Optical fibres are fabricated with high-quality composite glass or quartz fibres. Each fibre consists of a central core surrounded by a cladding. The refractive index of the material of the core is higher than that of the cladding.
Working: When a signal in the form of light is directed at one end of the fibre at a suitable angle,it undergoes repeated total internal reflections along the length of the fibre and finally emerges at the other end.
Since light undergoes total internal reflection at each stage,there is no appreciable loss in the intensity of the light signal. Optical fibres are designed such that light reflected at one side of the inner surface strikes the other at an angle larger than the critical angle. Even if the fibre is bent,light can easily travel along its length. Thus,an optical fibre acts as an optical pipe.
Applications: $A$ bundle of optical fibres can be used for several purposes. $A$ device that converts one form of energy into another is called a transducer. Optical fibres are extensively used for transmitting and receiving electrical signals,which are converted to light by suitable transducers. They are also used as a 'light pipe' to facilitate visual examination of internal organs like the esophagus,stomach,and intestines. Additionally,they are used in decorative lamps where light travels from the bottom of each fibre and appears at the tip as a dot of light.

(N/A) Total internal reflection occurs when the following two conditions are met:
$1$. The light must travel from a denser medium to a rarer medium.
$2$. The angle of incidence $(i)$ in the denser medium must be greater than the critical angle $(c)$ for the pair of media involved $(i > c)$.

(N/A) Total internal reflection is the phenomenon in which a light ray traveling from a denser medium to a rarer medium is reflected back into the same denser medium when the angle of incidence exceeds the critical angle for the pair of media.
Conditions for total internal reflection:
$1$. The light must travel from an optically denser medium to an optically rarer medium.
$2$. The angle of incidence $(i)$ must be greater than the critical angle $(C)$ for the given pair of media.

(N/A) The critical angle is defined as the angle of incidence in a denser medium for which the angle of refraction in the rarer medium is $90^{\circ}$.
When light travels from a denser medium to a rarer medium,as the angle of incidence increases,the angle of refraction also increases.
At a specific angle of incidence,the refracted ray grazes the interface between the two media,making the angle of refraction $90^{\circ}$. This specific angle of incidence is known as the critical angle ($C$ or $i_c$).

(A) An optical fibre works on the principle of Total Internal Reflection $(TIR)$.
For $TIR$ to occur, light must travel from a denser medium to a rarer medium.
Therefore, the refractive index of the core $(\mu_1)$ must be greater than the refractive index of the cladding $(\mu_2)$, i.e., $\mu_1 > \mu_2$.

(N/A) Optical fibres are extensively used in the medical field for a technique known as endoscopy.
$1$. An endoscope is a medical instrument consisting of a bundle of optical fibres that can be inserted into the body to view internal organs.
$2$. The principle behind this is $Total Internal Reflection$ $(TIR)$.
$3$. One bundle of fibres carries light into the body to illuminate the internal area,while another bundle transmits the reflected light back to the doctor's eyepiece or a camera.
$4$. This allows doctors to perform minimally invasive surgeries and examine internal structures like the stomach,intestines,or lungs without making large incisions.

(D) Suppose the required minimum diameter of the given disc is $d$. For light rays emanating from the point-like object $O$ and incident on the water surface from inside,if the angle of incidence $i \geq C$,then the object $O$ will not be seen by an observer outside the water due to total internal reflection. (Where $C$ is the critical angle for water to air).
Assume the angle $i$ in the figure is equal to $C$.
According to the formula for the critical angle:
$\sin C = \frac{1}{\mu}$
Since $i = C$,we have $\sin i = \frac{1}{\mu}$.
From the geometry of the figure:
$\tan i = \frac{d/2}{h}$
$\therefore \frac{d}{2} = h \tan i$
$\therefore d = 2h \tan i$ ... $(1)$
Now,since $\sin i = \frac{1}{\mu}$,we find $\cos i$:
$\cos i = \sqrt{1 - \sin^2 i} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$
Therefore,$\tan i = \frac{\sin i}{\cos i} = \frac{1/\mu}{\sqrt{\mu^2 - 1}/\mu} = \frac{1}{\sqrt{\mu^2 - 1}}$.
Substituting this into equation $(1)$:
$d = \frac{2h}{\sqrt{\mu^2 - 1}}$.

(N/A) As shown in the figure,consider a light ray $\overrightarrow{PQ}$ incident on the surface $AB$ of a denser transparent medium with angle of incidence $i$. After refraction,the ray $\overrightarrow{QR}$ is incident at point $R$ on the surface $AC$ of the rarer medium with angle of incidence $\phi$. For the light to be guided without escaping,it must undergo total internal reflection $(TIR)$ at every interface. Thus,we require $\phi \geq C$,where $C$ is the critical angle.
From the geometry,$\phi + r = 90^{\circ}$,so $\phi = 90^{\circ} - r$.
The condition for $TIR$ is $\phi \geq C$,which implies $\sin \phi \geq \sin C$.
Substituting $\phi = 90^{\circ} - r$,we get $\sin(90^{\circ} - r) \geq \sin C$,which simplifies to $\cos r \geq \frac{1}{\mu}$ (since $\sin C = \frac{1}{\mu}$).
Squaring both sides,$\cos^2 r \geq \frac{1}{\mu^2}$,or $1 - \sin^2 r \geq \frac{1}{\mu^2}$.
From Snell's law at point $Q$,$\sin i = \mu \sin r$,so $\sin r = \frac{\sin i}{\mu}$.
Substituting this,$1 - \frac{\sin^2 i}{\mu^2} \geq \frac{1}{\mu^2}$.
Multiplying by $\mu^2$,we get $\mu^2 - \sin^2 i \geq 1$,or $\mu^2 \geq 1 + \sin^2 i$.
Since the maximum value of $\sin^2 i$ is $1$ (at $i = 90^{\circ}$),the condition must hold for all $i$,so $\mu^2 \geq 1 + 1 = 2$,which means $\mu \geq \sqrt{2}$.
Thus,for $\mu \geq \sqrt{2}$,light will always undergo $TIR$ and be guided along the medium.

(C) The relationship between the critical angle $C$ and the refractive index $\mu$ is given by $\sin C = \frac{1}{\mu}$.
Given $C = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\mu}$.
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,it follows that $\frac{1}{\sqrt{2}} = \frac{1}{\mu}$,which implies $\mu = \sqrt{2}$.
The velocity of light in a medium $v$ is related to the speed of light in vacuum $c$ by $v = \frac{c}{\mu}$.
Substituting the values,$v = \frac{3 \times 10^{8}}{\sqrt{2}} \; m/s$.

(B) Assuming that the right-angled prism is an isosceles prism,the other two angles are $45^{\circ}$ each.
$\Rightarrow$ Each incident ray strikes the face $PR$ at an angle of incidence $i = 45^{\circ}$.
$\Rightarrow$ $A$ ray will emerge from the face $PR$ if the angle of incidence $i$ is less than the critical angle $\theta_{C}$ for that specific wavelength.
$\Rightarrow$ The critical angle is given by $\theta_{C} = \sin^{-1}\left(\frac{1}{\mu}\right)$.
$\Rightarrow$ For the red ray: $\mu_{R} = 1.27$. $\theta_{C,R} = \sin^{-1}\left(\frac{1}{1.27}\right) \approx 51.94^{\circ}$. Since $45^{\circ} < 51.94^{\circ}$,the red ray will emerge.
$\Rightarrow$ For the green ray: $\mu_{G} = 1.42$. $\theta_{C,G} = \sin^{-1}\left(\frac{1}{1.42}\right) \approx 44.76^{\circ}$. Since $45^{\circ} > 44.76^{\circ}$,the green ray will undergo total internal reflection and will not emerge.
$\Rightarrow$ For the blue ray: $\mu_{B} = 1.49$. $\theta_{C,B} = \sin^{-1}\left(\frac{1}{1.49}\right) \approx 42.15^{\circ}$. Since $45^{\circ} > 42.15^{\circ}$,the blue ray will undergo total internal reflection and will not emerge.
$\Rightarrow$ Therefore,only the red ray will emerge out of the face $PR$.

(A) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $r = i$.
The reflected and refracted rays make an angle of $90^{\circ}$ with each other. Since the sum of angles on a straight line is $180^{\circ}$,we have $r + 90^{\circ} + r^{\prime} = 180^{\circ}$.
Substituting $r = i$,we get $i + 90^{\circ} + r^{\prime} = 180^{\circ}$,which implies $r^{\prime} = 90^{\circ} - i$.
Using Snell's Law at the interface,$n_1 \sin i = n_2 \sin r^{\prime}$,where $n_1$ is the refractive index of the denser medium and $n_2$ is the refractive index of the rarer medium.
Substituting $r^{\prime} = 90^{\circ} - i$,we get $n_1 \sin i = n_2 \sin(90^{\circ} - i) = n_2 \cos i$.
Therefore,$\frac{n_2}{n_1} = \frac{\sin i}{\cos i} = \tan i$.
The critical angle $C$ is defined by $\sin C = \frac{n_2}{n_1}$.
Since $\frac{n_2}{n_1} = \tan i$ and $i = r$,we have $\sin C = \tan r$.
Thus,$C = \sin^{-1}(\tan r)$.

(C) The critical angle $\theta_{C}$ for the diamond-air interface is given by $\sin \theta_{C} = \frac{n_{air}}{n_{diamond}} = \frac{1}{2.42} \approx 0.4132$.
Calculating the critical angle: $\theta_{C} = \arcsin(0.4132) \approx 24.41^{\circ}$.
The angle of incidence is $\theta_{i} = 30^{\circ}$.
Since $\theta_{i} > \theta_{C}$ $(30^{\circ} > 24.41^{\circ})$,the condition for total internal reflection is satisfied.
Therefore,the light ray will undergo total internal reflection and will not refract into the air.

(B) For total internal reflection $(TIR)$ to occur at point $B$,the angle of incidence $\theta^{\prime \prime}$ must be greater than or equal to the critical angle $C$. The condition for grazing emergence at point $B$ is $\sin \theta^{\prime \prime} = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.
From the geometry of the triangle,we have $\theta^{\prime} = 90^{\circ} - \theta^{\prime \prime}$.
Applying Snell's Law at point $A$ on the top surface:
$1 \times \sin \theta = \mu \times \sin \theta^{\prime}$
$\sin \theta = \frac{4}{3} \times \sin(90^{\circ} - \theta^{\prime \prime})$
$\sin \theta = \frac{4}{3} \times \cos \theta^{\prime \prime}$
Since $\sin \theta^{\prime \prime} = \frac{3}{4}$,we find $\cos \theta^{\prime \prime} = \sqrt{1 - \sin^2 \theta^{\prime \prime}} = \sqrt{1 - (3/4)^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$.
Substituting this into the equation for $\sin \theta$:
$\sin \theta = \frac{4}{3} \times \frac{\sqrt{7}}{4} = \frac{\sqrt{7}}{3}$
$\theta = \sin ^{-1} \left( \frac{\sqrt{7}}{3} \right)$.

(B) Method $(i)$:
Using Snell's law,$n_1 \sin i = n_2 \sin r$.
Given $n_1 = 1$ (air),$n_2 = \sqrt{3}$ (glass),and $i = 60^{\circ}$.
$1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r$
$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$
$\sin r = \frac{1}{2} \implies r = 30^{\circ}$.
The angle between the reflected ray and the normal is $i = 60^{\circ}$.
The angle between the refracted ray and the normal is $r = 30^{\circ}$.
The angle between the reflected ray and the refracted ray is $180^{\circ} - (i + r) = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Method $(ii)$:
Since the angle of incidence $i = 60^{\circ}$ satisfies the condition for Brewster's angle,where $\tan i_p = \mu = \sqrt{3}$,it implies $i_p = 60^{\circ}$.
At Brewster's angle,the reflected and refracted rays are perpendicular to each other,so the angle between them is $90^{\circ}$.

(A) The refractive index $\mu$ is inversely proportional to the speed of light $v$ in a medium,given by $\mu = \frac{c}{v}$,where $c$ is the speed of light in vacuum.
The critical angle $i_c$ for light traveling from a denser medium to a rarer medium is given by the formula $\sin i_c = \frac{\mu_R}{\mu_D}$,where $\mu_R$ is the refractive index of the rarer medium and $\mu_D$ is the refractive index of the denser medium.
Since $\mu \propto \frac{1}{v}$,we have $\frac{\mu_R}{\mu_D} = \frac{v_D}{v_R}$.
Given $v_A = 1.5 \times 10^8 \ m/s$ and $v_B = 2.0 \times 10^8 \ m/s$,medium $A$ is the denser medium $(D)$ and medium $B$ is the rarer medium $(R)$.
Therefore,$\sin i_c = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{2.0 \times 10^8} = \frac{1.5}{2.0} = \frac{3}{4} = 0.750$.
Thus,$i_c = \sin^{-1}(0.750)$.

(D) For total internal reflection to occur,the angle of incidence $i$ must be greater than the critical angle $\theta_{C}$.
The refractive index of a medium is given by $\mu = \sqrt{\mu_{r} \epsilon_{r}}$.
For the given medium: $\mu_{D} = \sqrt{1 \times 4} = 2$.
For air: $\mu_{R} = 1$.
The critical angle $\theta_{C}$ is defined as $\sin \theta_{C} = \frac{\mu_{R}}{\mu_{D}} = \frac{1}{2}$.
Therefore,$\theta_{C} = 30^{\circ}$.
For total internal reflection,$i > \theta_{C}$,which means $i > 30^{\circ}$.
Among the given options,$60^{\circ}$ is the only value greater than $30^{\circ}$ that satisfies the condition for total internal reflection.

(C) Let $C$ be the critical angle.
From the geometry of the problem,the radius $r$ of the circular area on the water surface is given by $\tan C = \frac{r}{h}$,where $h = \sqrt{7} \; m$.
Thus,$r = h \tan C$.
We know that $\sin C = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.
Using the trigonometric identity $\tan C = \frac{\sin C}{\sqrt{1 - \sin^2 C}}$,we get:
$\tan C = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7/16}} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$.
Substituting the values,$r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3 \; m$.
The area of the surface is $A = \pi r^2 = \pi (3)^2 = 9 \pi \; m^2$.
Comparing this with $x \pi \; m^2$,we get $x = 9$.

(D) The speed of light in medium $B$ $(v_B)$ is $1.5 \times 10^{10} \, cm/s$ and in medium $A$ $(v_A)$ is $2.0 \times 10^{10} \, cm/s$.
Since $v_B < v_A$,medium $B$ is denser than medium $A$.
The critical angle $i_c$ for light traveling from a denser medium to a rarer medium is given by $\sin i_c = \frac{n_r}{n_d} = \frac{v_d}{v_r}$.
Here,$v_d = v_B = 1.5 \times 10^{10} \, cm/s$ and $v_r = v_A = 2.0 \times 10^{10} \, cm/s$.
$\sin i_c = \frac{1.5 \times 10^{10}}{2.0 \times 10^{10}} = \frac{1.5}{2.0} = \frac{3}{4}$.
Therefore,$i_c = \sin^{-1}\left(\frac{3}{4}\right)$.
For total internal reflection $(TIR)$ to occur,the angle of incidence $\theta$ must be greater than the critical angle $i_c$.
Thus,$\theta > \sin^{-1}\left(\frac{3}{4}\right)$.

(A) The speed of light in a medium is given by $V = \frac{c}{n}$,where $c$ is the speed of light in vacuum and $n$ is the refractive index.
Since $n = \frac{c}{V}$,the refractive index is inversely proportional to the speed of light.
For total internal reflection to occur,light must travel from a denser medium $(M_{1})$ to a rarer medium $(M_{2})$.
The condition for the critical angle $i_{c}$ is $n_{1} \sin i_{c} = n_{2} \sin 90^{\circ}$.
Thus,$\sin i_{c} = \frac{n_{2}}{n_{1}} = \frac{V_{1}}{V_{2}}$.
Given $V_{1} = 1.5 \times 10^{8} \text{ m/s}$ and $V_{2} = 2.0 \times 10^{8} \text{ m/s}$.
$\sin i_{c} = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} = \frac{1.5}{2.0} = \frac{3}{4}$.
If $\sin i_{c} = \frac{3}{4}$,then the opposite side is $3$ and the hypotenuse is $4$.
The adjacent side is $\sqrt{4^{2} - 3^{2}} = \sqrt{16 - 9} = \sqrt{7}$.
Therefore,$\tan i_{c} = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{\sqrt{7}}$.
Hence,$i_{c} = \tan^{-1}\left(\frac{3}{\sqrt{7}}\right)$.

(A) Given,$n_{L}=C_{A} n_{A}+(1-C_{A}) n_{B}$.
Here,$n_{A}=1.5$ and $n_{B}=1.3$.
Substituting these values,we get:
$n_{L} = C_{A}(1.5) + (1-C_{A})(1.3) = 1.3 + 0.2 C_{A}$.
Using Snell's law at the prism-liquid interface for the critical angle $\theta_{C}$:
$n_{p} \sin \theta_{C} = n_{L} \sin 90^{\circ}$.
Since $n_{p} = 1.5$,we have:
$\sin \theta_{C} = \frac{n_{L}}{1.5} = \frac{1.3 + 0.2 C_{A}}{1.5}$.
$\theta_{C} = \sin^{-1} \left( \frac{1.3 + 0.2 C_{A}}{1.5} \right)$.
At $C_{A} = 0$,$\theta_{C} = \sin^{-1} \left( \frac{1.3}{1.5} \right) = \sin^{-1} \left( \frac{13}{15} \right) \approx 60^{\circ}$.
At $C_{A} = 1$,$\theta_{C} = \sin^{-1} \left( \frac{1.5}{1.5} \right) = \sin^{-1}(1) = 90^{\circ}$.
As $C_{A}$ increases from $0$ to $1$,$\theta_{C}$ increases from $60^{\circ}$ to $90^{\circ}$. The function $\theta_{C} = \sin^{-1}(f(C_{A}))$ is a non-linear increasing function. Graph $(A)$ correctly represents this variation.

(D) From Snell's law,$n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}$.
For $\theta_{1} = \theta_{C}$,the angle of refraction $\theta_{2} = 90^{\circ}$,so $\sin \theta_{2} = 1$ and $\cos \theta_{2} = \sqrt{1 - \sin^{2} \theta_{2}} = 0$. Thus,option $(c)$ is true.
For $\theta_{1} > \theta_{C}$,$\sin \theta_{2} = \frac{n_{1}}{n_{2}} \sin \theta_{1} > 1$. Since $\sin \theta_{2} > 1$,$\cos \theta_{2} = \sqrt{1 - \sin^{2} \theta_{2}}$ becomes imaginary. Thus,option $(b)$ is true.
For any angle of incidence $\theta_{1}$,the reflected ray follows the law of reflection,so $\theta_{1} = \theta_{3}$. Thus,option $(a)$ is true.
Regarding option $(d)$,$\theta_{3}$ is the angle of reflection,which is always real and equal to $\theta_{1}$. Therefore,$\cos \theta_{3}$ is always real,making the statement in option $(d)$ false.

(B) The dielectric slab is bounded by the planes $x=0, x=a$ and $y=0, y=b$. $A$ light ray incident at $y=0$ at an angle $\theta$ refracts into the dielectric at an angle $r$ with the normal.
For the ray to reach $y=b$,it must undergo Total Internal Reflection $(TIR)$ at the boundary $x=a$. The angle of incidence at the boundary $x=a$ is $(90^{\circ}-r)$.
For $TIR$ to occur at $x=a$,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin C = 1/n$. Thus,$\sin(90^{\circ}-r) \geq 1/n$,which simplifies to $\cos r \geq 1/n$,or $n \geq 1/\cos r$.
From Snell's law at the boundary $y=0$,$n = \sin \theta / \sin r$. Since $\theta$ can range from $0^{\circ}$ to $90^{\circ}$,the limiting case occurs at $\theta = 90^{\circ}$,giving $n = 1/\sin r$,or $\sin r = 1/n$.
Using the identity $\sin^2 r + \cos^2 r = 1$,we have $(1/n)^2 + (1/n)^2 = 1$,which leads to $2/n^2 = 1$,or $n^2 = 2$. Therefore,the minimum refractive index is $n = \sqrt{2}$.

(D) Let the angle of incidence be $i$ and the angle of refraction be $r$. According to Snell's law at the first surface:
$\frac{\sin i}{\sin r} = \mu = \sqrt{3} \implies \sin i = \sqrt{3} \sin r \quad \dots(1)$
In the triangle formed by the center of the sphere and the two points of refraction/reflection,the triangle is isosceles because two sides are radii of the sphere. The angle of deviation at the first surface is $(i - r)$. The angle of reflection at the second surface is also $r$.
Since the incident ray is parallel to the emergent ray,the total deviation $\delta$ must be $180^{\circ}$.
The deviation at the first refraction is $(i - r)$.
The deviation at the internal reflection is $(180^{\circ} - 2r)$.
The deviation at the second refraction is $(i - r)$.
Total deviation $\delta = (i - r) + (180^{\circ} - 2r) + (i - r) = 180^{\circ}$.
$2i - 4r + 180^{\circ} = 180^{\circ} \implies 2i = 4r \implies i = 2r$.
Substitute $i = 2r$ into equation $(1)$:
$\sin(2r) = \sqrt{3} \sin r$
$2 \sin r \cos r = \sqrt{3} \sin r$
Since $\sin r \neq 0$,we have $\cos r = \frac{\sqrt{3}}{2}$.
Thus,$r = 30^{\circ}$.
Since $i = 2r$,we get $i = 2 \times 30^{\circ} = 60^{\circ}$.
Therefore,the angle of incidence is $60^{\circ}$.

(A) Let $R$ be the radius of the circular region through which outside objects are visible.
Let $\theta$ be the critical angle for the water-air interface.
Then,$\sin \theta = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.
Now,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}}$.
Substituting the value of $\sin \theta$:
$\tan \theta = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7/16}} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$.
From the geometry of the problem,the diver's eyes are at a height $h$ from the bottom,so the distance from the water surface is $(H-h)$.
Thus,$\tan \theta = \frac{R}{H-h}$.
Equating the two expressions for $\tan \theta$:
$\frac{R}{H-h} = \frac{3}{\sqrt{7}}$.
Therefore,$R = \frac{3(H-h)}{\sqrt{7}}$.

(B) For the light beam to pass through the material without undergoing total internal reflection at the interface,the condition for the critical angle must be satisfied at the boundary between the material and the glass.
From Snell's Law at the interface,$n_1 \sin i = n_2 \sin r$. For the limiting case of total internal reflection,the angle of refraction $r = 90^{\circ}$.
Here,the refractive index of glass is $n_1 = 1.5$ and the minimum refractive index of the material is $n_2 = 1.2$.
Thus,$\sin i_{\max} = \frac{n_2}{n_1} = \frac{1.2}{1.5} = 0.8$.
Therefore,$i_{\max} = \sin^{-1}(0.8) = 53.1^{\circ}$.

(B) For the light source to be invisible from above,the light rays emitted from the source must undergo total internal reflection at the water-air interface.
Let $r$ be the radius of the opaque disc. The light rays reaching the edge of the disc must strike the surface at an angle equal to the critical angle $i_c$.
From the geometry of the problem,$\tan i_c = \frac{r}{h}$.
We know that for total internal reflection,$\sin i_c = \frac{1}{\mu}$.
Using the trigonometric identity $\tan i_c = \frac{\sin i_c}{\sqrt{1-\sin^2 i_c}}$,we get:
$\tan i_c = \frac{1/\mu}{\sqrt{1-(1/\mu)^2}} = \frac{1/\mu}{\sqrt{(\mu^2-1)/\mu^2}} = \frac{1}{\sqrt{\mu^2-1}}$.
Equating the two expressions for $\tan i_c$:
$\frac{r}{h} = \frac{1}{\sqrt{\mu^2-1}}$
Therefore,$r = \frac{h}{\sqrt{\mu^2-1}}$.

(D) The correct option is $(D)$.
$1$. At the first interface (between medium $1$ and medium $2$),the light ray bends towards the normal. According to Snell's law,when a ray bends towards the normal,it travels from a rarer medium to a denser medium. Therefore,$\mu_2 > \mu_1$.
$2$. At the second interface (between medium $2$ and medium $3$),the light ray undergoes total internal reflection,as no light enters the third medium. Total internal reflection occurs only when light travels from a denser medium to a rarer medium. Therefore,$\mu_2 > \mu_3$.
$3$. Since the light is totally internally reflected at the second interface,the angle of incidence must be greater than the critical angle for the pair of media $(2, 3)$. This implies that medium $2$ is optically denser than medium $3$. Combining these observations,we have $\mu_2 > \mu_1$ and $\mu_2 > \mu_3$. Furthermore,since the ray is bent towards the normal at the first interface and undergoes total internal reflection at the second,the refractive index $\mu_1$ must be greater than $\mu_3$ to satisfy the geometry of the path. Thus,the correct order is $\mu_3 < \mu_1 < \mu_2$.

(A) The formation of a primary rainbow involves the following steps:
$1$. When sunlight enters a spherical water droplet,it undergoes refraction and dispersion.
$2$. The light then strikes the inner surface of the droplet and undergoes total internal reflection.
$3$. Finally,the light exits the droplet,undergoing refraction again as it passes from water into air.
Therefore,the correct sequence of processes is: refraction,total internal reflection,and refraction.
This corresponds to option $(A)$.

(C) The prism is isosceles with $\angle Q = 120^{\circ}$. Since the sum of angles in a triangle is $180^{\circ}$,$\angle P = \angle R = (180^{\circ} - 120^{\circ}) / 2 = 30^{\circ}$.
The rays are incident normally on face $PQ$,so they enter the prism without deviation and strike face $PR$ at an angle of incidence $i = 30^{\circ}$.
For a ray to emerge from face $PR$,it must satisfy the condition for refraction,which is $n \sin i < 1$ (where $n$ is the refractive index of the prism).
If $n \sin i \geq 1$,total internal reflection occurs at face $PR$.
Here,$i = 30^{\circ}$,so $\sin i = 0.5$.
The condition for total internal reflection is $n \times 0.5 \geq 1$,which means $n \geq 2$.
For rays $1$ $(n=1.85)$ and $2$ $(n=1.95)$,$n < 2$,so they refract out through face $PR$.
For rays $3$ $(n=2.05)$ and $4$ $(n=2.15)$,$n > 2$,so they undergo total internal reflection at face $PR$ and are directed towards face $QR$.
Since the angle of incidence at face $QR$ for these reflected rays is less than the critical angle,rays $3$ and $4$ emerge from face $QR$.

(C) The light beam will escape the water surface only when the angle of incidence $i$ is less than or equal to the critical angle $i_c$.
The critical angle $i_c$ for water-air interface is given by:
$\sin i_c = \frac{1}{n} = \frac{1}{1.33} \approx 0.7519$
$i_c = \sin^{-1}(0.7519) \approx 48.75^{\circ}$
The laser rotates in a vertical plane. It will emit light out of the water surface when its angle with the vertical is between $-i_c$ and $+i_c$. Thus, the total angular range for which the light escapes is $2i_c$.
The angular velocity of the laser is $\omega = \frac{2\pi}{T} = \frac{2\pi}{60} \, \text{rad/s}$.
The time $t$ for which the light escapes is given by:
$t = \frac{\text{angular range}}{\omega} = \frac{2i_c}{\omega}$
Converting $i_c$ to radians:
$i_c = 48.75^{\circ} \times \frac{\pi}{180^{\circ}} \approx 0.8508 \, \text{rad}$
$t = \frac{2 \times 0.8508}{\frac{2\pi}{60}} = \frac{0.8508 \times 60}{\pi} \approx 16.25 \, s$.
Using the approximation $i_c \approx 50^{\circ}$ as suggested by the provided solution structure:
$t = \frac{2 \times (50^{\circ} \times \frac{\pi}{180^{\circ}})}{\frac{2\pi}{60}} = \frac{100}{180} \times 60 = \frac{100}{3} \approx 33.33 \, s$ is incorrect based on the arc logic. The correct calculation for $i_c \approx 50^{\circ}$ is $t = \frac{2 \times 50}{360} \times 60 = \frac{100}{6} = 16.67 \, s$.

(A) As $\theta$ increases,the angle of incidence $i = 90^\circ - \theta$ decreases.
Initially,for small $\theta$,the angle of incidence $i$ is greater than the critical angle $i_c$,so total internal reflection $(TIR)$ occurs. The reflected ray hits the screen at a positive height $x = h \tan \theta$,where $h$ is the distance of the source from the surface. As $\theta$ increases,$x$ increases.
When $\theta$ reaches the value such that $i = i_c$,the ray grazes the surface.
For $\theta$ greater than this critical value,$i < i_c$,and the ray refracts out of the glass slab. The refracted ray hits the screen below the horizontal axis,making $x$ negative. As $\theta$ increases further,the angle of refraction increases,causing the spot to move further down,making $x$ more negative. Thus,the correct plot is $A$.