Critical Angle and Total Internal Reflection Questions in English

(C) The refractive index $\mu$ of a medium is given by the ratio of the wavelength in vacuum (or air) $\lambda_a$ to the wavelength in the medium $\lambda_m$:
$\mu = \frac{\lambda_a}{\lambda_m} = \frac{6000 \ Å}{4000 \ Å} = \frac{6}{4} = \frac{3}{2}$.
The relationship between the critical angle $C$ and the refractive index $\mu$ is given by:
$\sin C = \frac{1}{\mu}$.
Substituting the value of $\mu$:
$\sin C = \frac{1}{3/2} = \frac{2}{3}$.
Therefore,the critical angle is:
$C = \sin ^{-1}\left(\frac{2}{3}\right)$.

(A) The rays are incident normally on the first surface of the prism,so they pass undeviated and strike the hypotenuse at an angle of incidence $i = 45^{\circ}$.
For total internal reflection to occur at the second surface,the angle of incidence must be greater than the critical angle $(i > C)$.
Given critical angles are $C_{red} = 46^{\circ}$,$C_{green} = 44^{\circ}$,and $C_{blue} = 43^{\circ}$.
For red light: $i = 45^{\circ} < C_{red} = 46^{\circ}$. Thus,red light will emerge from the prism.
For green light: $i = 45^{\circ} > C_{green} = 44^{\circ}$. Thus,green light will undergo total internal reflection.
For blue light: $i = 45^{\circ} > C_{blue} = 43^{\circ}$. Thus,blue light will undergo total internal reflection.
Since green and blue light undergo total internal reflection while red light emerges,the arrangement separates red color from blue and green colors.

(B) ray of light incident normally on one face of a right-angled isosceles prism passes undeviated into the prism.
When it reaches the hypotenuse,the angle of incidence $i$ is $45^{\circ}$.
Since the ray grazes the hypotenuse,the angle of refraction is $90^{\circ}$.
Thus,the angle of incidence is equal to the critical angle $C$,so $C = 45^{\circ}$.
The refractive index $\mu$ is given by $\mu = \frac{1}{\sin C}$.
Substituting the value of $C$,we get $\mu = \frac{1}{\sin 45^{\circ}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.
Therefore,$\mu \approx 1.414$.

(C) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $r$,so $i = r$.
Given that the reflected and refracted rays are mutually perpendicular,the sum of the angle of reflection $r$,the angle between them $(90^{\circ})$,and the angle of refraction $r'$ is $180^{\circ}$.
Thus,$r + 90^{\circ} + r' = 180^{\circ}$.
Since $r = i$,we have $i + 90^{\circ} + r' = 180^{\circ}$,which implies $r' = 90^{\circ} - i$.
Applying Snell's Law at the interface: $\mu_r \sin i = \mu_d \sin r'$.
Substituting $r' = 90^{\circ} - i$,we get $\mu_r \sin i = \mu_d \sin(90^{\circ} - i) = \mu_d \cos i$.
Therefore,$\frac{\mu_d}{\mu_r} = \frac{\sin i}{\cos i} = \tan i$.
For the critical angle $C$,we know that $\sin C = \frac{\mu_r}{\mu_d}$.
Thus,$\sin C = \frac{1}{\tan i} = \cot i$.
Hence,the critical angle is $C = \sin^{-1}(\cot i)$.

(B) The refractive index of glass with respect to air is $\mu = 1.5$.
The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.
$C = \sin^{-1}(0.667) \approx 41.8^{\circ} \approx 42^{\circ}$.
Since the angle of incidence $i = 50^{\circ}$ is greater than the critical angle $C = 42^{\circ}$,the light ray undergoes total internal reflection.
In total internal reflection,the angle of reflection $r$ is equal to the angle of incidence $i$,so $r = 50^{\circ}$.
The deviation $\delta$ is the angle between the original path of the incident ray and the reflected ray.
From the geometry of the reflection,the deviation is $\delta = 180^{\circ} - (i + r) = 180^{\circ} - (50^{\circ} + 50^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

(A) Let the refractive index of the glass slab be $\mu$. The angle of incidence at point $A$ is $i = 45^\circ$. Let the angle of refraction at $A$ be $r$. By Snell's Law,$1 \cdot \sin(45^\circ) = \mu \cdot \sin(r)$,so $\sin(r) = \frac{1}{\mu\sqrt{2}}$.
At point $B$,the angle of incidence is $r' = 90^\circ - r$. For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,where $\sin(C) = \frac{1}{\mu}$.
Thus,$\sin(90^\circ - r) \ge \frac{1}{\mu}$,which implies $\cos(r) \ge \frac{1}{\mu}$.
Squaring both sides,$\cos^2(r) \ge \frac{1}{\mu^2}$,so $1 - \sin^2(r) \ge \frac{1}{\mu^2}$.
Substituting $\sin^2(r) = \frac{1}{2\mu^2}$,we get $1 - \frac{1}{2\mu^2} \ge \frac{1}{\mu^2}$,which simplifies to $1 \ge \frac{3}{2\mu^2}$.
Thus,$\mu^2 \ge \frac{3}{2}$,or $\mu \ge \sqrt{\frac{3}{2}}$. The minimum refractive index is $\sqrt{\frac{3}{2}}$.

(C) When a light ray travels from a denser to a rarer medium,it undergoes refraction. The angle of deviation $\delta$ for refraction is given by $\delta = |r - i|$,where $i$ is the angle of incidence and $r$ is the angle of refraction.
Since $r > i$,$\delta = r - i$.
As $i$ increases,$r$ also increases. The maximum value of $i$ is the critical angle $C$,at which point $r = 90^\circ$ or $\frac{\pi}{2}$ radians.
Thus,the maximum deviation for refraction is $\delta_{\max} = \frac{\pi}{2} - C$.
However,if the angle of incidence $i$ exceeds $C$,the ray undergoes total internal reflection. In this case,the angle of reflection is equal to the angle of incidence $(r = i)$.
The deviation for reflection is $\delta = \pi - 2i$.
To find the maximum deviation for reflection,we consider the range $C < i < \frac{\pi}{2}$. As $i$ approaches $C$ from the denser side,$\delta$ approaches $\pi - 2C$. This is the maximum possible deviation for the ray in this scenario.

(D) The critical angle $i_c$ is given by the formula $\sin(i_c) = \frac{1}{n}$,where $n$ is the refractive index of the material with respect to air.
For diamond,the refractive index $n \approx 2.42$.
Therefore,$\sin(i_c) = \frac{1}{2.42} \approx 0.413$.
Taking the inverse sine,$i_c = \arcsin(0.413) \approx 24.4^{\circ}$.

(C) The critical angle $C$ is given by the formula $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the medium.
Since the refractive index $n$ is inversely proportional to the wavelength $\lambda$ (Cauchy's relation: $n \approx A + \frac{B}{\lambda^2}$),as the wavelength decreases,the refractive index increases.
For red light,the wavelength is $\lambda_1$. For yellow light,the wavelength is $\lambda_2$. Since $\lambda_1 > \lambda_2$,the refractive index for red light $(n_r)$ is less than the refractive index for yellow light $(n_y)$.
Therefore,$n_r < n_y$.
Since $\sin C = \frac{1}{n}$,a higher refractive index results in a smaller critical angle.
Thus,the critical angle for yellow light will be less than the critical angle for red light $(\theta)$.

(A) The critical angle $(i_c)$ is the angle of incidence in a denser medium for which the angle of refraction in the rarer medium is $90^{\circ}$.
Given,refractive index of the denser medium $\mu_1 = 2$ and refractive index of the rarer medium $\mu_2 = 1$.
The formula for the critical angle is given by $\sin(i_c) = \frac{\mu_2}{\mu_1}$.
Substituting the values,we get $\sin(i_c) = \frac{1}{2}$.
Therefore,$i_c = \sin^{-1}(0.5) = 30^{\circ}$.

(C) The critical angle $i_c$ is defined as the angle of incidence for which the angle of refraction is $90^{\circ}$ when light travels from a denser medium to a rarer medium.
Given,refractive index of the first medium $\mu_1 = 2$ and refractive index of the second medium $\mu_2 = \sqrt{3}$.
The formula for the critical angle is given by $\sin i_c = \frac{\mu_2}{\mu_1}$.
Substituting the values,we get $\sin i_c = \frac{\sqrt{3}}{2}$.
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have $i_c = 60^{\circ}$.

(C) The refractive index $\mu$ of a diamond is very high (approximately $2.42$).
Since the critical angle $\theta_c$ is given by the relation $\sin \theta_c = \frac{1}{\mu}$,a high refractive index results in a very small critical angle.
When light enters a well-cut diamond,it strikes the internal surfaces at angles greater than the critical angle.
This leads to multiple total internal reflections within the diamond.
Consequently,the light is trapped and reflected multiple times,making the diamond appear bright.

(A) The refractive index of glass with respect to water is given by the ratio of their absolute refractive indices (or refractive indices with respect to air).
Given: Refractive index of water,$\mu_w = \frac{4}{3}$.
Refractive index of glass,$\mu_g = \frac{5}{3}$.
The relative refractive index of glass with respect to water is $\mu = \frac{\mu_g}{\mu_w} = \frac{5/3}{4/3} = \frac{5}{4}$.
The critical angle $C$ is defined by the relation $\sin C = \frac{1}{\mu}$.
Substituting the value of $\mu$,we get $\sin C = \frac{1}{5/4} = \frac{4}{5}$.
Therefore,the critical angle $C = \sin ^{-1}\left(\frac{4}{5}\right)$.

(A) The refractive index of the first medium is $\mu_1 = 2$ and the refractive index of the second medium is $\mu_2 = \sqrt{2}$.
Total internal reflection occurs when the light travels from a denser medium to a rarer medium and the angle of incidence $i$ is greater than the critical angle $i_C$.
The critical angle $i_C$ is given by the formula $\sin i_C = \frac{\mu_2}{\mu_1}$.
Substituting the values,we get $\sin i_C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i_C = 45^{\circ}$.
For total internal reflection to occur,the angle of incidence must be greater than the critical angle,i.e.,$i > 45^{\circ}$.

(A) The prism is a right-angled triangle with angles $30^{\circ}, 60^{\circ}, 90^{\circ}$.
When a ray of light is incident normally on face $AB$,it passes undeviated into the prism.
It then strikes the hypotenuse face at an angle of incidence $i = 60^{\circ}$.
For total internal reflection $(TIR)$ to occur at the hypotenuse face,the angle of incidence must be greater than the critical angle $C$,i.e.,$i > C$.
Thus,$\sin(i) > \sin(C)$.
Given $i = 60^{\circ}$,we have $\sin(60^{\circ}) > \frac{\mu_l}{\mu_{\text{glass}}}$.
$\frac{\sqrt{3}}{2} > \frac{\mu_l}{3/2}$.
$\frac{\sqrt{3}}{2} > \frac{2 \mu_l}{3}$.
$\mu_l < \frac{3 \sqrt{3}}{4}$.
Therefore,the refractive index of the liquid must be less than $\frac{3 \sqrt{3}}{4}$.

(A) According to Snell's Law,the refractive index of the liquid with respect to air $(n_{la})$ is given by:
$n_{la} = \frac{\sin \theta}{\sin \alpha}$
By definition,the critical angle $\theta_c$ is the angle of incidence in the denser medium (liquid) for which the angle of refraction in the rarer medium (air) is $90^{\circ}$.
The refractive index of air with respect to the liquid $(n_{al})$ is:
$n_{al} = \frac{1}{n_{la}} = \frac{\sin \theta_c}{\sin 90^{\circ}}$
Since $\sin 90^{\circ} = 1$,we have:
$\sin \theta_c = \frac{1}{n_{la}}$
Substituting the value of $n_{la}$:
$\sin \theta_c = \frac{1}{\frac{\sin \theta}{\sin \alpha}} = \frac{\sin \alpha}{\sin \theta}$

(D) The critical angle $\theta_{c}$ is the angle of incidence for which the angle of refraction is $90^{\circ}$.
It is given by the formula:
$\sin \theta_{c} = \frac{n_2}{n_1}$
Given $n_1 = 2$ and $n_2 = \sqrt{3}$,we substitute these values into the formula:
$\sin \theta_{c} = \frac{\sqrt{3}}{2}$
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we have:
$\theta_{c} = 60^{\circ}$

(C) Given: Refractive indices $\mu_1 = 1.6$,$\mu_2 = 1.8$,$\mu_3 = 1.3$.
At the interface between medium $2$ and $3$,the ray is incident at the critical angle $C$. Therefore,$\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.
Let $r$ be the angle of refraction in medium $2$. Since the ray is incident at the critical angle at the interface of $2$ and $3$,the angle of refraction $r$ at the first interface is equal to the critical angle $C$ (i.e.,$r = C$).
Applying Snell's Law at the interface between medium $1$ and $2$:
$\mu_1 \sin \theta = \mu_2 \sin r$
Since $r = C$,we have $\sin r = \sin C = \frac{1.3}{1.8}$.
Substituting the values:
$1.6 \times \sin \theta = 1.8 \times \left(\frac{1.3}{1.8}\right)$
$1.6 \times \sin \theta = 1.3$
$\sin \theta = \frac{1.3}{1.6} = \frac{13}{16}$
$\theta = \sin ^{-1}\left(\frac{13}{16}\right)$

(C) The acceptance angle $\theta_a$ is the maximum angle of incidence at the entrance of the optical fibre such that the light undergoes total internal reflection at the core-cladding interface.
For an optical fibre with core refractive index $\mu_1 = 1.5$ and cladding refractive index $\mu_2 = 1.414$,the acceptance angle $\theta_a$ is given by the formula:
$\sin \theta_a = \sqrt{\mu_1^2 - \mu_2^2}$
Substituting the given values:
$\sin \theta_a = \sqrt{(1.5)^2 - (1.414)^2}$
$\sin \theta_a = \sqrt{2.25 - 1.999396} \approx \sqrt{0.2506} \approx 0.5006$
$\theta_a = \sin^{-1}(0.5006) \approx 30^{\circ}$
Thus,the range of the incident angle with the axis of the optical fibre for total internal reflection is $0^{\circ}$ to $30^{\circ}$.
Therefore,option $C$ is correct.

(B) The assertion $(A)$ is true because optical fibres work on the principle of total internal reflection $(TIR)$,which occurs when light travels from a denser medium (core) to a rarer medium (cladding) at an angle of incidence greater than the critical angle.
The reason $(R)$ states that the refractive index of the core is greater than that of air. While it is true that the core has a higher refractive index than air,the condition for $TIR$ in an optical fibre is that the refractive index of the core $(n_1)$ must be greater than the refractive index of the cladding $(n_2)$,not necessarily air.
Therefore,while both statements are factually correct,the reason $(R)$ does not explain why $TIR$ occurs at the core-clad interface. The correct explanation would involve the relationship between the core and the cladding refractive indices.

(A) Optical fibres are extensively used in modern communication systems because they offer high bandwidth and low signal loss.
They are physically small,lightweight,and flexible,making them easy to install.
Since light signals are confined within the fibre due to total internal reflection,they are immune to electromagnetic interference,which is a significant advantage over copper wires.
Therefore,both the assertion and the reason are true,and the reason correctly explains why optical fibres are widely used.

(D) For total internal reflection to occur at the hypotenuse,the angle of incidence $i$ must be greater than or equal to the critical angle $C$.
From the geometry of the prism,the light ray travels parallel to the base $BC$. The angle between the incident ray and the normal to the hypotenuse is the angle of incidence $i$.
In the right-angled triangle,if the base angle is $\theta$,the angle of incidence $i$ at the hypotenuse is $90^{\circ} - \theta$.
For total internal reflection,$i \geq C$,where $\sin C = \frac{1}{\mu}$.
Thus,$90^{\circ} - \theta \geq C \Rightarrow \theta \leq 90^{\circ} - C$.
To find the maximum value of $\theta$,we set $\theta = 90^{\circ} - C$.
Taking the cosine on both sides: $\cos \theta = \cos(90^{\circ} - C) = \sin C$.
Since $\sin C = \frac{1}{\mu}$,we have $\cos \theta = \frac{1}{\mu}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{1}{\mu}\right)$.

(A) The given situation is shown in the figure. The light ray enters the prism perpendicular to the face $AB$,so it passes undeviated into the prism and strikes the face $AC$.
In the triangle formed by the prism,the angle of incidence $i$ at the face $AC$ is $i = 90^{\circ} - \theta$.
For total internal reflection to occur at face $AC$,the angle of incidence $i$ must be greater than or equal to the critical angle $i_c$.
Thus,for the largest value of $\theta$,we set $i = i_c$.
Using Snell's law at the interface $AC$ for the critical angle condition:
$\sin(i_c) = \frac{\mu_{\text{liquid}}}{\mu_{\text{glass}}}$
Substituting $i = 90^{\circ} - \theta$ and the given refractive indices:
$\sin(90^{\circ} - \theta) = \frac{1.2}{1.5}$
$\cos(\theta) = \frac{12}{15} = \frac{4}{5} = 0.8$
Therefore,$\theta = \cos^{-1}(0.8)$.

(C) Given,radius $R=0.05 \ m = 5 \times 10^{-2} \ m$.
Refractive index $n=1.5$.
The critical angle $c$ is given by $\sin c = \frac{1}{n} = \frac{1}{1.5} = \frac{2}{3}$.
From the geometry of the problem,the ray that strikes the curved surface at the critical angle $c$ will emerge tangent to the surface and hit the table at a distance $x$ from the edge of the cylinder.
In the right-angled triangle formed by the radius $R$ and the distance $(R+x)$,the angle at the center is $c$.
Thus,$\cos c = \frac{R}{R+x}$.
Since $\sin c = \frac{2}{3}$,we have $\cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.
Equating the two expressions for $\cos c$:
$\frac{\sqrt{5}}{3} = \frac{R}{R+x}$.
$\sqrt{5}(R+x) = 3R$.
$\sqrt{5}x = 3R - \sqrt{5}R = R(3 - \sqrt{5})$.
$x = R \frac{3 - \sqrt{5}}{\sqrt{5}} = R \left( \frac{3}{\sqrt{5}} - 1 \right) = R \left( \frac{3\sqrt{5}}{5} - 1 \right)$.
Wait,re-evaluating the geometry: The distance from the center to the point where the ray hits the table is $d = \frac{R}{\cos c}$.
So,$R+x = \frac{R}{\cos c} = \frac{R}{\sqrt{5}/3} = \frac{3R}{\sqrt{5}} = \frac{3\sqrt{5}R}{5}$.
$x = \frac{3\sqrt{5}R}{5} - R = R \left( \frac{3\sqrt{5}-5}{5} \right)$.
Given $R = 5 \times 10^{-2} \ m$,we get $x = 5 \times 10^{-2} \left( \frac{3\sqrt{5}-5}{5} \right) = (3\sqrt{5}-5) \times 10^{-2} \ m$.

(C) When a diver is inside the water,they see the outside world through a cone of light due to the phenomenon of total internal reflection.
The semi-vertical angle of this cone is equal to the critical angle $c$ for the water-air interface.
The formula for the critical angle is given by $\sin c = \frac{1}{\mu}$.
Given the refractive index of water $\mu = \frac{4}{3}$,we have:
$\sin c = \frac{1}{4/3} = \frac{3}{4}$.
Therefore,the semi-vertical angle is $c = \sin ^{-1}\left(\frac{3}{4}\right)$.

(B) When a fish is at a depth $h$ below the surface of water,it sees the outside world through a circular window due to total internal reflection.
The radius $r$ of this circle is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$.
Given: depth $h = 12 \ m$ and refractive index $\mu = 4/3$.
Substituting the values into the formula:
$r = \frac{12}{\sqrt{(4/3)^2 - 1}}$
$r = \frac{12}{\sqrt{16/9 - 1}}$
$r = \frac{12}{\sqrt{7/9}}$
$r = \frac{12 \times 3}{\sqrt{7}}$
$r = \frac{36}{\sqrt{7}} \ m$.

(C) For total internal reflection to occur,light must travel from a denser medium to a rarer medium. Here,the speed of light in medium $A$ $(v_A = 2.4 \times 10^8 \ m/s)$ is less than in medium $B$ $(v_B = 2.7 \times 10^8 \ m/s)$,so medium $A$ is denser.
At the critical angle $c$,the angle of refraction is $90^\circ$. According to Snell's Law:
$\mu_A \sin c = \mu_B \sin 90^\circ$
Since $\mu = \frac{c_{light}}{v}$,we have $\frac{c_{light}}{v_A} \sin c = \frac{c_{light}}{v_B} \times 1$
$\sin c = \frac{v_A}{v_B} = \frac{2.4 \times 10^8}{2.7 \times 10^8} = \frac{24}{27} = \frac{8}{9}$
Now,using the trigonometric identity $\tan c = \frac{\sin c}{\sqrt{1 - \sin^2 c}}$:
$\tan c = \frac{8/9}{\sqrt{1 - (8/9)^2}} = \frac{8/9}{\sqrt{1 - 64/81}} = \frac{8/9}{\sqrt{17/81}} = \frac{8}{\sqrt{17}}$
Therefore,$c = \tan^{-1}(\frac{8}{\sqrt{17}})$.