Refraction Through Prism Questions in English

(C) Let $n_1$ be the refractive index of the medium outside the first surface, $n_2$ be the refractive index of the prism, and $n_3$ be the refractive index of the medium outside the second surface.
For a ray to bend towards the normal, it must travel from a rarer to a denser medium $(n_{incident} < n_{refracted})$.
For a ray to bend away from the normal, it must travel from a denser to a rarer medium $(n_{incident} > n_{refracted})$.
In case $(a)$, the ray bends towards the normal at the first surface $(n_1 < n_2)$ and grazes the second surface, implying total internal reflection or critical angle condition $(n_2 > n_3)$. Thus, $n_2 > n_1$ and $n_2 > n_3$ is correct.
In case $(b)$, the ray enters undeviated at the first surface $(n_1 = n_2)$ and bends away from the normal at the second surface $(n_2 > n_3)$. This is correct.
In case $(c)$, the ray bends towards the normal at the first surface $(n_1 < n_2)$ and bends away from the normal at the second surface $(n_2 > n_3)$. The statement says $n_2 < n_1$ and $n_2 = n_3$, which contradicts the observed bending. Thus, $(c)$ is the wrong statement.

(D) For a prism,the angle of deviation is given by $\delta = (i + e) - A$.
In the position of minimum deviation,the angle of incidence is equal to the angle of emergence,i.e.,$i = e$.
Under this condition,the light ray inside the prism travels parallel to the base,and the angle of refraction $r$ is given by $r_1 = r_2 = r$.
Since $A = r_1 + r_2$,we have $A = 2r$.
Given the prism angle $A = 60^{\circ}$,the angle of refraction is $r = \frac{A}{2} = \frac{60^{\circ}}{2} = 30^{\circ}$.
This condition is independent of the wavelength or colour of the light.
Therefore,the angle of refraction is $30^{\circ}$ for both red and violet colours.

(B) For an equilateral prism,the angle of prism or refracting angle is $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_{m}$ is equal to the refracting angle $A$,so $\delta_{m} = A = 60^{\circ}$.
The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$.
Substituting the values,we get $\mu = \frac{\sin((60^{\circ} + 60^{\circ})/2)}{\sin(60^{\circ}/2)}$.
$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we have $\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(C) The dispersive power $\omega$ of a prism is given by $\omega = \frac{\delta_{B} - \delta_{R}}{\delta_{y}}$,where $\delta_{B}$ and $\delta_{R}$ are the deviations for blue and red rays,and $\delta_{y}$ is the mean deviation,calculated as $\delta_{y} = \frac{\delta_{B} + \delta_{R}}{2}$.
For the first prism:
$\delta_{B1} = 12^{\circ}$,$\delta_{R1} = 8^{\circ}$.
$\delta_{y1} = \frac{12^{\circ} + 8^{\circ}}{2} = 10^{\circ}$.
$\omega_{1} = \frac{12^{\circ} - 8^{\circ}}{10^{\circ}} = \frac{4}{10} = \frac{2}{5}$.
For the second prism:
$\delta_{B2} = 14^{\circ}$,$\delta_{R2} = 10^{\circ}$.
$\delta_{y2} = \frac{14^{\circ} + 10^{\circ}}{2} = 12^{\circ}$.
$\omega_{2} = \frac{14^{\circ} - 10^{\circ}}{12^{\circ}} = \frac{4}{12} = \frac{1}{3}$.
The ratio of dispersive powers is $\frac{\omega_{1}}{\omega_{2}} = \frac{2/5}{1/3} = \frac{2}{5} \times 3 = \frac{6}{5}$.

(B) The formula for the refractive index of the prism material $(\mu)$ with respect to the surrounding liquid is given by: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$, where $A$ is the refracting angle and $\delta_m$ is the angle of minimum deviation.
Given $A = 60^{\circ}$ and $\delta_m = 30^{\circ}$, we have:
$\mu = \frac{\sin((60^{\circ} + 30^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
The critical angle $(C)$ is related to the refractive index by the formula: $\sin(C) = 1/\mu$.
Substituting the value of $\mu$: $\sin(C) = 1/\sqrt{2}$.
Therefore, $C = 45^{\circ}$.

(D) For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.
When the light ray inside the prism moves parallel to the base,the ray undergoes minimum deviation,which implies that the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r_1$ is equal to $r_2$.
Since $r_1 + r_2 = A$,we have $2r = 60^{\circ}$,which gives $r = 30^{\circ}$.
Using Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.
Given $\mu = \sqrt{3}$ and $r = 30^{\circ}$,we have $\sqrt{3} = \frac{\sin i}{\sin 30^{\circ}}$.
Since $\sin 30^{\circ} = 0.5$,we get $\sin i = \sqrt{3} \times 0.5 = \frac{\sqrt{3}}{2}$.
Therefore,$i = 60^{\circ}$.

(D) For a light ray to retrace its path after reflection from the silvered second face, it must strike the second face normally (at an angle of $90^{\circ}$ to the surface).
Let $i = 60^{\circ}$ be the angle of incidence on the first face, $r_1$ be the angle of refraction at the first face, and $A = 30^{\circ}$ be the prism angle.
Since the ray strikes the second face normally, the angle of refraction at the second face is $r_2 = 0^{\circ}$.
From the geometry of the prism, we know that $A = r_1 + r_2$.
Substituting the values, $30^{\circ} = r_1 + 0^{\circ}$, so $r_1 = 30^{\circ}$.
Using Snell's Law at the first face: $\mu = \frac{\sin(i)}{\sin(r_1)}$.
Substituting the values: $\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.
Thus, the refractive index of the material of the prism is $\sqrt{3}$.

(D) In a prism,let the angle of the prism be $A$.
When a light ray enters the first face,it refracts at an angle $r_1$.
When it reaches the second face,it strikes at an angle $r_2$ with respect to the normal.
From the geometry of the quadrilateral formed by the two normals and the two refracting faces of the prism,the sum of the angle of the prism $A$ and the angle between the two normals is $180^{\circ}$.
Also,in the triangle formed by the refracted ray inside the prism and the two refracting faces,the sum of the angles is $180^{\circ}$,which gives $A + r_1 + r_2 = 180^{\circ}$ (considering the interior angles of the triangle).
Thus,the relationship between the angle of the prism and the angles of refraction is $A = r_1 + r_2$.

(A) When a ray is incident normally on one surface of the prism,it passes undeviated into the prism. Let the prism be $ABC$ with refracting angle $A = 30^{\circ}$.
At the second surface $AB$,the angle of incidence $i$ is equal to the refracting angle of the prism,so $i = 30^{\circ}$.
Using Snell's law at the second surface: $\mu \sin i = 1 \sin e$,where $e$ is the angle of emergence.
$1.5 \times \sin 30^{\circ} = \sin e$
$1.5 \times 0.5 = \sin e$
$\sin e = 0.75$
Given $\sin 48^{\circ} 36^{\prime} = 0.75$,therefore $e = 48^{\circ} 36^{\prime}$.
The angle of deviation $\delta$ for a ray incident normally is given by $\delta = e - i$.
$\delta = 48^{\circ} 36^{\prime} - 30^{\circ} = 18^{\circ} 36^{\prime}$.

(D) Given,the angle of incidence $i = 0^{\circ}$.
For an equilateral prism,the prism angle $A = 60^{\circ}$.
At the first surface,since the ray is incident normally,the angle of refraction $r_1 = 0^{\circ}$.
Using the relation $r_1 + r_2 = A$,we get $0^{\circ} + r_2 = 60^{\circ}$,so $r_2 = 60^{\circ}$.
The critical angle $C$ for the prism is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$,which means $C = 45^{\circ}$.
Since the angle of incidence at the second surface $r_2 = 60^{\circ}$ is greater than the critical angle $C = 45^{\circ}$,the light ray undergoes total internal reflection at the second surface.
After reflection,the ray strikes the third surface. The angle of incidence at the third surface is $r_3 = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$ (from the geometry of the triangle). Since $60^{\circ} > 45^{\circ}$,it undergoes another total internal reflection.
Finally,the ray emerges from the first surface normally. The total deviation $\delta$ is the angle between the incident ray and the emergent ray. Since the ray emerges parallel to the incident ray but in the opposite direction,the deviation is $180^{\circ} - 60^{\circ} = 120^{\circ}$.

(C) For a light ray to be refracted from the second surface,the angle of incidence at the second surface $(r_2)$ must be less than the critical angle $(C)$.
The condition for refraction at the second surface is $r_2 < C$.
The critical angle $C$ is given by $\sin C = 1/\mu = 1/\sqrt{7/3} = \sqrt{3/7}$.
Thus,$\sin r_2 < \sqrt{3/7}$.
For a prism,$r_1 + r_2 = A = 60^{\circ}$,so $r_2 = 60^{\circ} - r_1$.
Substituting this,$\sin(60^{\circ} - r_1) < \sqrt{3/7}$.
To find the minimum angle of incidence $(i)$,we need the maximum value of $r_1$. Since $r_2$ must be less than $C$,$r_1$ must be greater than $60^{\circ} - C$.
Using $\sin C = \sqrt{3/7} \approx 0.6546$,we find $C \approx 40.89^{\circ}$.
So,$r_1 > 60^{\circ} - 40.89^{\circ} = 19.11^{\circ}$.
Using Snell's law at the first surface: $\sin i = \mu \sin r_1 = \sqrt{7/3} \sin(19.11^{\circ}) \approx 1.5275 \times 0.3274 \approx 0.5$.
Thus,$i > 30^{\circ}$. The minimum angle is $30^{\circ}$.

(C) Given: Refractive index $\mu = \sqrt{3}$ and angle of minimum deviation $\delta_m = A$.
For a prism,the refractive index is given by the formula: $\mu = \frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}$.
Substituting the given values: $\sqrt{3} = \frac{\sin \frac{A+A}{2}}{\sin \frac{A}{2}}$.
$\sqrt{3} = \frac{\sin A}{\sin \frac{A}{2}}$.
Using the trigonometric identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we get: $\sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}}$.
$\sqrt{3} = 2 \cos \frac{A}{2}$.
$\cos \frac{A}{2} = \frac{\sqrt{3}}{2}$.
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,we have $\frac{A}{2} = 30^{\circ}$.
Therefore,$A = 60^{\circ}$.

(B) Given: Angle of incidence $i = 45^{\circ}$,Prism angle $A = 30^{\circ}$.
Since the ray retraces its path after reflection from the silvered face,it must strike the silvered face normally (at $90^{\circ}$).
In the triangle formed by the ray and the prism sides,the angle at the silvered face is $90^{\circ}$ and the prism angle is $30^{\circ}$.
Therefore,the angle of refraction $r$ at the first face is $r = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$ is incorrect based on the geometry shown. Looking at the triangle,the angle of refraction $r$ is $r = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Using Snell's Law: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}}$.
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) For a prism,at the condition of minimum deviation,the angle of refraction $r$ is related to the prism angle $A$ by $A = 2r$.
Given $r = 30^{\circ}$,we have $A = 2 \times 30^{\circ} = 60^{\circ}$.
The formula for the refractive index $\mu$ of the prism material is $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.
Substituting the given values $\delta_m = 30^{\circ}$ and $A = 60^{\circ}$:
$\mu = \frac{\sin((60^{\circ} + 30^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})}$.
Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we get $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) Let the prism angle be $A = 90^{\circ}$ and refractive index be $\mu = \sqrt{2}$.
For total internal reflection $(TIR)$ at the second face, the angle of incidence at the second face $r_2$ must be greater than or equal to the critical angle $C$.
The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$, which gives $C = 45^{\circ}$.
Thus, for $TIR$, $r_2 > 45^{\circ}$.
From the geometry of the prism, $r_1 + r_2 = A = 90^{\circ}$.
Substituting $r_2 > 45^{\circ}$, we get $r_1 < 90^{\circ} - 45^{\circ} = 45^{\circ}$.
Using Snell's law at the first face: $\sin i = \mu \sin r_1$.
Since $r_1 < 45^{\circ}$, $\sin r_1 < \sin 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore, $\sin i < \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$, which implies $i < 90^{\circ}$.
However, for the light to enter the prism and satisfy the condition, we look for the boundary condition where $r_2 = 45^{\circ}$, which implies $r_1 = 45^{\circ}$.
Then $\sin i = \sqrt{2} \sin 45^{\circ} = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1$, so $i = 90^{\circ}$.
Since the question asks for the condition for $TIR$, any angle of incidence $i < 90^{\circ}$ will result in $r_1 > 45^{\circ}$ which leads to $r_2 < 45^{\circ}$ (no $TIR$). Thus, the boundary is at $i = 90^{\circ}$.

(A) For a small-angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = (\mu - 1)A$,where $\mu$ is the refractive index and $A$ is the angle of the prism.
The angle of dispersion (angular dispersion) $\theta$ is the difference between the angles of deviation for violet and red light: $\theta = \delta_v - \delta_r$.
Substituting the formula for deviation: $\theta = (\mu_v - 1)A - (\mu_r - 1)A = (\mu_v - \mu_r)A$.
Given values: $\mu_v = 1.52$,$\mu_r = 1.48$,and $A = 6^{\circ}$.
Calculation: $\theta = (1.52 - 1.48) \times 6^{\circ} = 0.04 \times 6^{\circ} = 0.24^{\circ}$.
Therefore,the angle of dispersion is $0.24^{\circ}$.

(C) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_m = A = 60^{\circ}$.
The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$.
Substituting the values,we get $n = \frac{\sin((60^{\circ} + 60^{\circ}) / 2)}{\sin(60^{\circ} / 2)}$.
$n = \frac{\sin(120^{\circ} / 2)}{\sin(30^{\circ})} = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we have $n = \frac{\sqrt{3} / 2}{1 / 2} = \sqrt{3}$.
The value of $\sqrt{3}$ is approximately $1.732$.

(C) For a small-angle prism,the relationship between the angle of incidence $i$,the angle of refraction $r_1$ at the first face,the angle of refraction $r_2$ at the second face,and the angle of emergence $e$ is given by:
$A = r_1 + r_2$
Given that the ray emerges normally from the second face,the angle of emergence $e = 0^{\circ}$,which implies that $r_2 = 0^{\circ}$.
Since the refracting angle $A = 6^{\circ}$,we have $r_1 = A - r_2 = 6^{\circ} - 0^{\circ} = 6^{\circ}$.
According to Snell's Law at the first face,$n_1 \sin(i) = n_2 \sin(r_1)$,where $n_1 = 1$ (air) and $n_2 = n$ (prism material).
$1 \cdot \sin(9.3^{\circ}) = n \cdot \sin(6^{\circ})$.
Using the small-angle approximation $\sin(\theta) \approx \theta$ (in radians) or simply the ratio of angles for small values:
$n = \frac{\sin(9.3^{\circ})}{\sin(6^{\circ})} \approx \frac{9.3}{6} = 1.55$.
Thus,the refractive index of the material of the prism is $1.55$.

(D) For a prism,the relationship between the angles is given by $A + \delta = i + e$,where $A$ is the angle of the prism,$\delta$ is the angle of deviation,$i$ is the angle of incidence,and $e$ is the angle of emergence.
Given that the prism is equilateral,the angle of the prism $A = 60^{\circ}$.
The angle of incidence $i$ and the angle of emergence $e$ are both equal to $70 \%$ of the angle of the prism.
Therefore,$i = e = 0.70 \times 60^{\circ} = 42^{\circ}$.
Substituting these values into the formula $A + \delta = i + e$:
$60^{\circ} + \delta = 42^{\circ} + 42^{\circ}$
$60^{\circ} + \delta = 84^{\circ}$
$\delta = 84^{\circ} - 60^{\circ} = 24^{\circ}$.
Thus,the angle of deviation is $24^{\circ}$.

(A) For a thin prism,the angle of minimum deviation $\delta$ is given by the formula $\delta = A(\mu - 1)$,where $A$ is the prism angle and $\mu$ is the refractive index of the prism material relative to the surrounding medium.
$1$. When the prism is in air:
The refractive index of the prism with respect to air is $\mu_1 = \frac{3}{2}$.
Therefore,the angle of minimum deviation is $\delta_1 = A(\frac{3}{2} - 1) = A(\frac{1}{2}) = \frac{A}{2}$.
$2$. When the prism is in water:
The refractive index of water is $\mu_w = \frac{4}{3}$. The refractive index of the prism with respect to water is $\mu_2 = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
Therefore,the angle of minimum deviation is $\delta_2 = A(\frac{9}{8} - 1) = A(\frac{1}{8}) = \frac{A}{8}$.
$3$. The ratio of the angles of minimum deviation is:
$\frac{\delta_1}{\delta_2} = \frac{A/2}{A/8} = \frac{8}{2} = \frac{4}{1}$.
Thus,the ratio is $4: 1$.

(B) Given: Refractive index $\mu = 1.6$,Angle of minimum deviation $\delta_m = 4.2^{\circ}$.
For a small-angled prism,the relationship between the angle of deviation $\delta$,refractive index $\mu$,and prism angle $A$ is given by the formula: $\delta = (\mu - 1)A$.
Substituting the given values into the formula:
$4.2^{\circ} = (1.6 - 1)A$
$4.2^{\circ} = (0.6)A$
$A = \frac{4.2^{\circ}}{0.6} = 7^{\circ}$.
Therefore,the angle of the prism is $7^{\circ}$.

(B) For a light ray passing through a prism,the angle of deviation $\delta$ at the first surface is $\theta_1 = i - r_1$ and at the second surface is $\theta_2 = e - r_2$.
The total deviation $\delta$ is the sum of deviations at both surfaces:
$\delta = \theta_1 + \theta_2$
$\delta = (i - r_1) + (e - r_2)$
$\delta = (i + e) - (r_1 + r_2)$
Since the prism angle $A = r_1 + r_2$,we substitute this into the equation:
$\delta = i + e - A$

(C) For dispersion without deviation through a prism combination,the net deviation must be zero.
$\delta_{net} = \delta_1 - \delta_2 = 0$
$\Rightarrow \delta_1 = \delta_2$
For thin prisms,the deviation is given by $\delta = A(\mu - 1)$.
Therefore,$A_1(\mu_1 - 1) = A_2(\mu_2 - 1)$.
Given: $A_1 = 9^{\circ}$,$\mu_1 = 1.4$,and $\mu_2 = 1.6$.
Substituting the values:
$9^{\circ} \times (1.4 - 1) = A \times (1.6 - 1)$
$9^{\circ} \times 0.4 = A \times 0.6$
$A = \frac{9 \times 0.4}{0.6} = \frac{3.6}{0.6} = 6^{\circ}$.

(B) Let $\delta_m$ be the angle of minimum deviation and $A$ be the angle of the prism.
We know the formula for the refractive index $\mu$ of a prism is given by:
$\mu = \frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Given that $\mu = \sqrt{2}$ and $\delta_m = A$,we substitute these values into the formula:
$\sqrt{2} = \frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\sqrt{2} = \frac{\sin A}{\sin \left(\frac{A}{2}\right)}$
Using the trigonometric identity $\sin A = 2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)$,we get:
$\sqrt{2} = \frac{2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\sqrt{2} = 2 \cos \left(\frac{A}{2}\right)$
$\cos \left(\frac{A}{2}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$
Since $\cos(45^{\circ}) = \frac{1}{\sqrt{2}}$,we have:
$\frac{A}{2} = 45^{\circ}$
$A = 90^{\circ}$

(C) For dispersion without deviation,the net deviation produced by the combination must be zero.
$\delta - \delta^{\prime} = 0 \Rightarrow A(\mu - 1) - A^{\prime}(\mu^{\prime} - 1) = 0$
Where:
$A = 6^{\circ}$ (angle of the first prism)
$\mu = 1.5$ (refractive index of the first prism)
$A^{\prime} = ?$ (angle of the second prism)
$\mu^{\prime} = 1.75$ (refractive index of the second prism)
Substituting the values into the equation:
$A^{\prime}(\mu^{\prime} - 1) = A(\mu - 1)$
$A^{\prime}(1.75 - 1) = 6(1.5 - 1)$
$A^{\prime}(0.75) = 6(0.5)$
$A^{\prime} = \frac{6 \times 0.5}{0.75} = \frac{3}{0.75} = 4^{\circ}$
Thus,the angle of the second prism is $4^{\circ}$.

(B) Let the angle of the prism be $A$. Since the ray is incident normally on face $AB$,it enters the prism without deviation.
At the silvered face $AC$,the angle of incidence $i_1$ is equal to the prism angle $A$ (as the normal to $AC$ makes an angle $A$ with the normal to $AB$).
By the law of reflection,the angle of reflection is also $i_1 = A$.
In the triangle formed by the ray and the faces,the angle at the base $AB$ is $90^{\circ} - A$. Thus,the angle of incidence at the second reflection on face $AB$ is $i_2 = 90^{\circ} - (90^{\circ} - 2A) = 2A$.
Finally,the ray emerges from the base $BC$ normally. In the triangle near vertex $B$,the angles are $B$,$90^{\circ} - i_2$,and $90^{\circ}$. Since $AB = AC$,the base angles are $B = C = (180^{\circ} - A) / 2 = 90^{\circ} - A/2$.
From the geometry of the path,the sum of angles in the triangle formed by the ray inside the prism is $A + 2A + 2A = 180^{\circ}$.
$5A = 180^{\circ} \implies A = 36^{\circ}$.

(B) The formula for the refractive index $\mu$ of a prism in terms of the refracting angle $A$ and the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$
Given $\mu = \cot(A / 2) = \frac{\cos(A / 2)}{\sin(A / 2)}$,we substitute this into the formula:
$\frac{\cos(A / 2)}{\sin(A / 2)} = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$
Canceling $\sin(A / 2)$ from both sides,we get:
$\cos(A / 2) = \sin((A + \delta_m) / 2)$
Using the trigonometric identity $\cos \theta = \sin(90^{\circ} - \theta)$ or $\sin(\pi/2 - \theta)$:
$\sin(\pi/2 - A/2) = \sin((A + \delta_m) / 2)$
Equating the angles:
$\pi/2 - A/2 = (A + \delta_m) / 2$
Multiplying by $2$:
$\pi - A = A + \delta_m$
$\delta_m = \pi - 2A$

(C) Given: Refractive index $\mu = \sqrt{3}$ and angle of minimum deviation $\delta_m = A$.
The formula for the refractive index of a prism is given by:
$\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$
Substituting the given values:
$\sqrt{3} = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})}$
$\sqrt{3} = \frac{\sin(A)}{\sin(\frac{A}{2})}$
Using the trigonometric identity $\sin(A) = 2 \sin(\frac{A}{2}) \cos(\frac{A}{2})$:
$\sqrt{3} = \frac{2 \sin(\frac{A}{2}) \cos(\frac{A}{2})}{\sin(\frac{A}{2})}$
$\sqrt{3} = 2 \cos(\frac{A}{2})$
$\cos(\frac{A}{2}) = \frac{\sqrt{3}}{2}$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have:
$\frac{A}{2} = 30^{\circ}$
$A = 60^{\circ}$

(A) For two prisms combined to produce dispersion without deviation (achromatic combination),the net deviation must be zero.
Let the mean deviations produced by the two prisms be $d$ and $d^{\prime}$.
The condition for zero net deviation is given by $d + d^{\prime} = 0$.
However,the question asks for the condition of dispersion without deviation. The dispersive power $\omega$ is defined as $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_y$ is the mean deviation $(d)$.
Thus,the angular dispersion is $\theta = \omega d$.
For the net dispersion to be non-zero while the net deviation is zero,we require $d + d^{\prime} = 0$.
Given the standard form for such problems,the condition for zero deviation is $d + d^{\prime} = 0$. If the options provided are meant to represent the relationship between dispersive powers and deviations,the correct physical condition is $d + d^{\prime} = 0$. Since the provided options seem to contain typos,the most physically relevant form for zero deviation is $d + d^{\prime} = 0$.

(B) Given: Angle of incidence $i = 45^{\circ}$,Prism angle $A = 30^{\circ}$.
Since the ray retraces its path after reflection from the silvered face,it must strike the silvered face normally (at $90^{\circ}$).
In the triangle formed by the ray inside the prism,the angles are $A = 30^{\circ}$,the angle at the silvered face is $90^{\circ}$,and the angle of refraction $r$ at the first face.
The sum of angles in a triangle is $180^{\circ}$,so $r + 90^{\circ} + 30^{\circ} = 180^{\circ}$.
$r = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
Wait,looking at the geometry: the angle of refraction $r$ is the angle with the normal. The angle inside the prism at the first face is $90^{\circ} - r$.
Thus,$(90^{\circ} - r) + 90^{\circ} + 30^{\circ} = 180^{\circ}$.
$210^{\circ} - r = 180^{\circ} \implies r = 30^{\circ}$.
Using Snell's Law: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}}$.
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(B) Given: Refracting angle $A = 60^{\circ}$,Angle of minimum deviation $\delta_m = 30^{\circ}$.
Using the formula for the refractive index of a prism immersed in a liquid:
$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
$\mu = \frac{\sin \left( \frac{60^{\circ} + 30^{\circ}}{2} \right)}{\sin \left( \frac{60^{\circ}}{2} \right)} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}}$
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
The critical angle $C$ is given by $\sin C = \frac{1}{\mu}$.
$\sin C = \frac{1}{\sqrt{2}}$
$C = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45^{\circ}$.

(A, C) The critical angle $\theta_{C} = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$.
Applying Snell's law at the incident surface $S$ $(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2})$:
$1 \cdot \sin 45^{\circ} = \sqrt{2} \cdot \sin \theta \implies \sin \theta = \frac{1}{2} \implies \theta = 30^{\circ}$.
At point $P$,the angle of incidence is $90^{\circ} - 15^{\circ} = 75^{\circ}$. Since $75^{\circ} > 45^{\circ}$,total internal reflection $(TIR)$ occurs at $P$.
At point $Q$,the angle of incidence is $15^{\circ}$. Since $15^{\circ} < 45^{\circ}$,partial reflection and refraction occur at $Q$.
The ray emerging at $R$ is parallel to the incident ray at $S$ because the geometry of the prism and the path of the ray result in a net deviation of $0^{\circ}$ (or $360^{\circ}$),meaning they are parallel. Thus,statement $A$ and $C$ are correct.

(B) For a prism,the angle of deviation $\delta$ is given by the formula: $\delta = i + e - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is the prism angle.
From the graph,we observe that for a deviation $\delta = 30^{\circ}$,there are two possible angles of incidence: $i_1 = 15^{\circ}$ and $i_2 = 60^{\circ}$.
By the principle of reversibility of light,if $i = 15^{\circ}$,then $e = 60^{\circ}$,and if $i = 60^{\circ}$,then $e = 15^{\circ}$.
Substituting these values into the formula: $A = i + e - \delta$.
$A = 15^{\circ} + 60^{\circ} - 30^{\circ}$.
$A = 75^{\circ} - 30^{\circ} = 45^{\circ}$.
Therefore,the prism angle is $45^{\circ}$.

(D) For dispersion without deviation,the net deviation produced by the combination of two prisms must be zero.
Let the refractive indices of prisms $P_{1}$ and $P_{2}$ be $\mu$ and $\mu^{\prime}$ and their refracting angles be $A$ and $A^{\prime}$ respectively.
The condition for no deviation is given by $\delta + \delta^{\prime} = 0$,which implies $(\mu - 1)A = (\mu^{\prime} - 1)A^{\prime}$.
Given values are $\mu = 1.54$,$A = 4^{\circ}$,and $A^{\prime} = 3^{\circ}$.
Substituting these values into the equation: $(1.54 - 1) \times 4^{\circ} = (\mu^{\prime} - 1) \times 3^{\circ}$.
$0.54 \times 4 = (\mu^{\prime} - 1) \times 3$.
$2.16 = (\mu^{\prime} - 1) \times 3$.
$\mu^{\prime} - 1 = \frac{2.16}{3} = 0.72$.
$\mu^{\prime} = 0.72 + 1 = 1.72$.

(B) The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.
Given that the angle of minimum deviation $\delta_{\min}$ is equal to the refracting angle $A$,we substitute $\delta_{\min} = A$ into the formula.
$n = \frac{\sin((A + A)/2)}{\sin(A/2)} = \frac{\sin(A)}{\sin(A/2)}$.
Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get $n = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.
Since the refracting angle $A$ of a prism must satisfy $0 < A < 180^{\circ}$,the value of $\cos(A/2)$ lies between $\cos(90^{\circ}) = 0$ and $\cos(0^{\circ}) = 1$.
Therefore,$0 < \cos(A/2) < 1$.
Multiplying by $2$,we get $0 < 2 \cos(A/2) < 2$,which implies $0 < n < 2$.
However,for a physical prism,$A$ is typically small or less than $180^{\circ}$,and $n > 1$. Thus,$1 < n < 2$.

(A) For the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r$ is given by $r = \frac{A}{2}$,where $A$ is the prism angle.
At the exit surface,the light ray strikes the interface between the prism (refractive index $n$) and the coating (refractive index $n/2$).
For the condition of the critical angle $\theta_{c}$,the angle of refraction $r$ must be equal to $\theta_{c}$.
According to Snell's law at the exit surface: $n \sin(r) = (n/2) \sin(90^{\circ})$.
Since $\sin(90^{\circ}) = 1$,we have $n \sin(r) = n/2$.
Dividing both sides by $n$,we get $\sin(r) = 1/2$.
Since $r = A/2$,we have $\sin(A/2) = 1/2$.
Therefore,$A/2 = 30^{\circ}$,which gives $A = 60^{\circ}$.

(A) Let the prism be $PDC$ with base $DC$. The incident ray is parallel to the base $DC$.
For grazing emergence at the second surface,the angle of refraction $r_2$ is equal to the critical angle $C$.
Given $\mu = \sqrt{2}$,we have $\sin r_2 = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$,which gives $r_2 = 45^{\circ}$.
Since the incident ray is parallel to the base,the angle of incidence $i$ at the first surface is equal to the base angle $45^{\circ}$.
Applying Snell's Law at the first surface: $1 \times \sin(45^{\circ}) = \sqrt{2} \times \sin(r_1)$.
$\frac{1}{\sqrt{2}} = \sqrt{2} \sin(r_1) \implies \sin(r_1) = \frac{1}{2} \implies r_1 = 30^{\circ}$.
The angle of the prism $A$ at the top vertex is $A = r_1 + r_2 = 30^{\circ} + 45^{\circ} = 75^{\circ}$.
In the triangle formed by the prism,the sum of angles is $180^{\circ}$. Thus,$45^{\circ} + \theta + A = 180^{\circ}$.
$45^{\circ} + \theta + 75^{\circ} = 180^{\circ} \implies \theta + 120^{\circ} = 180^{\circ} \implies \theta = 60^{\circ}$.

(C) For a thin prism,the deviation produced is given by $\delta = (\mu - 1)A$.
For dispersion without deviation,the net deviation must be zero,i.e.,$\delta_{net} = \delta_{1} + \delta_{2} = 0$.
Since the prisms are combined to produce dispersion without deviation,they must be placed in opposite orientations,so $(\mu_{1} - 1)A_{1} + (\mu_{2} - 1)A_{2} = 0$.
Taking the magnitude,we have $(\mu_{1} - 1)A_{1} = -(\mu_{2} - 1)A_{2}$.
Substituting the given values: $(1.72 - 1) \times 5^{\circ} = -(1.9 - 1) \times A_{2}$.
$0.72 \times 5^{\circ} = -0.9 \times A_{2}$.
The negative sign indicates the orientation of the second prism. The magnitude of the angle $A_{2}$ is:
$A_{2} = \frac{0.72 \times 5^{\circ}}{0.9} = \frac{3.6^{\circ}}{0.9} = 4^{\circ}$.

(A) For an equilateral prism,the angle of the prism $ A = 60^{\circ} $.
When the emergent ray grazes the surface,the angle of emergence $ e = 90^{\circ} $.
According to Snell's law at the second surface: $ \mu \sin(r_2) = 1 \cdot \sin(e) $.
Substituting the values: $ \sqrt{2} \cdot \sin(r_2) = 1 \cdot \sin(90^{\circ}) = 1 $.
$ \sin(r_2) = \frac{1}{\sqrt{2}} $.
Therefore,$ r_2 = 45^{\circ} $.
Using the relation $ A = r_1 + r_2 $,we find the angle of refraction at the first surface:
$ r_1 = A - r_2 = 60^{\circ} - 45^{\circ} = 15^{\circ} $.

(D) For the condition of minimum deviation in an equilateral prism,the angle of incidence $i$ is equal to the angle of emergence $e$.
The relationship is given by $i = e = \frac{A + \delta_m}{2}$.
For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
Given the angle of minimum deviation $\delta_m = 46^{\circ}$.
Substituting these values into the formula:
$i = \frac{60^{\circ} + 46^{\circ}}{2} = \frac{106^{\circ}}{2} = 53^{\circ}$.
Therefore,the angle of incidence is $53^{\circ}$.

(A) $1$. From the figure,the incident ray strikes the first surface of the prism normally. Therefore,the angle of incidence $i = 0^\circ$,and the ray passes undeviated into the prism.
$2$. The prism is a right-angled triangle with one angle $60^\circ$. The angle at the top vertex is $180^\circ - 90^\circ - 60^\circ = 30^\circ$.
$3$. Inside the prism,the ray hits the second surface (the hypotenuse). The angle of incidence at this surface $(r_2)$ is the angle between the normal and the ray. Since the ray is perpendicular to the first surface,the angle of incidence at the second surface is $r_2 = 30^\circ$.
$4$. Applying Snell's law at the second surface: $n_1 \sin(r_2) = n_2 \sin(e)$,where $n_1 = \sqrt{3}$,$n_2 = 1$,and $r_2 = 30^\circ$.
$5$. $\sqrt{3} \sin(30^\circ) = 1 \times \sin(e)$.
$6$. $\sqrt{3} \times \frac{1}{2} = \sin(e) \implies \sin(e) = \frac{\sqrt{3}}{2}$.
$7$. Therefore,$e = 60^\circ$.

(B) For a thin prism,the angle of minimum deviation is given by $\delta_m = (\mu - 1)A$,where $\mu$ is the refractive index and $A$ is the prism angle.
For a symmetric prism,the angle of incidence $i$ at minimum deviation is given by $i = \frac{A + \delta_m}{2}$.
Substituting $\delta_m = (\mu - 1)A$ into the expression for $i$:
$i = \frac{A + (\mu - 1)A}{2} = \frac{A + \mu A - A}{2} = \frac{\mu A}{2}$.
Now,the ratio of the angle of incidence $i$ to the angle of minimum deviation $\delta_m$ is:
$\frac{i}{\delta_m} = \frac{\mu A / 2}{(\mu - 1)A} = \frac{\mu}{2(\mu - 1)}$.
Given $\mu = 1.5$,we have:
$\frac{i}{\delta_m} = \frac{1.5}{2(1.5 - 1)} = \frac{1.5}{2(0.5)} = \frac{1.5}{1} = \frac{3}{2}$.
Thus,the ratio is $3 : 2$.