Refraction Through Prism Questions in English

(A) Ordinary glass prisms,such as $Crown$ or $Flint$ glass,absorb infrared radiation and are therefore unsuitable for studying the infrared spectrum.
$Rock-salt$ $(NaCl)$ crystals are transparent to infrared radiation,allowing them to pass through without significant absorption.
Therefore,a $Rock-salt$ prism is used to obtain and study the spectrum of infrared radiation.

(B) For an equilateral prism,the angle of the prism $A = 60^o$.
Given that the incident ray is perpendicular to the first refracting surface,the angle of incidence $i = 0^o$.
Therefore,the angle of refraction at the first surface $r_1 = 0^o$.
The angle of incidence at the second surface is $r_2 = A - r_1 = 60^o - 0^o = 60^o$.
The critical angle $C$ is given as $45^o$.
Since the angle of incidence at the second surface $(60^o)$ is greater than the critical angle $(45^o)$,total internal reflection occurs at the second surface.
The reflected ray then strikes the third surface (the base) at an angle of incidence of $30^o$ (calculated from the geometry of the triangle).
Since $30^o < 45^o$,the ray refracts out of the third surface.
Thus,the ray is totally reflected on the second surface and emerges out from the third surface.

(C) The deviation produced by a thin prism is given by the formula $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the material of the prism and $A$ is the angle of the prism.
Since the refractive index of glass is lowest for red light $(\mu_R < \mu_V)$,the deviation $\delta$ is directly proportional to $(\mu - 1)$.
Therefore,the deviation is minimum for the red ray ($\delta_R$ is least).

(C) Dispersion is the phenomenon of splitting of white light into its constituent colours.
In a medium like glass, the refractive index $(n)$ depends on the wavelength $(\lambda)$ of the light, a property known as dispersion.
The speed of light in a medium is given by $v = c/n$.
Since different colours (wavelengths) have different refractive indices in the flint glass, they travel at different speeds $(v)$ within the prism.
This difference in speed leads to different angles of deviation for each colour, causing the light to disperse.
Therefore, the correct reason is that light of different colours travels with different speeds in the glass medium.

(A) Given: Refractive index $\mu = 1.5$,Prism angle $A = 30^\circ$.
Since the ray is incident normally on one surface,the angle of incidence $i_1 = 0^\circ$,which implies the angle of refraction $r_1 = 0^\circ$.
Using the relation $r_1 + r_2 = A$,we get $0^\circ + r_2 = 30^\circ$,so $r_2 = 30^\circ$.
Applying Snell's Law at the second surface: $\mu \sin r_2 = 1 \cdot \sin e$,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 $\sin 48^\circ 36' = 0.75$,therefore $e = 48^\circ 36'$.
The deviation $\delta$ for a prism is given by $\delta = i_1 + e - A$.
Since $i_1 = 0^\circ$,$\delta = e - A = 48^\circ 36' - 30^\circ = 18^\circ 36'$.
Thus,the correct option is $A$.

(D) For minimum deviation,the angle of incidence $i$ is related to the angle of prism $A$ and the angle of minimum deviation $\delta_m$ by the relation $i = \frac{A + \delta_m}{2}$.
Given that the angle of incidence $i = 45^o$ and the refractive index $\mu = \sqrt{2}$.
The formula for the refractive index of a prism is $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(A/2)}$.
Substituting $i = \frac{A + \delta_m}{2} = 45^o$,we get $\mu = \frac{\sin(45^o)}{\sin(A/2)}$.
Substituting the values: $\sqrt{2} = \frac{1/\sqrt{2}}{\sin(A/2)}$.
This simplifies to $\sin(A/2) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$.
Therefore,$A/2 = 30^o$,which gives $A = 60^o$.

(D) The dispersive power $\omega$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.
Mathematically,$\omega = \frac{\delta_v - \delta_r}{\delta}$,where $\delta_v - \delta_r$ is the angular dispersion and $\delta$ is the mean deviation.
Therefore,the angular dispersion $\theta = \delta_v - \delta_r = \omega \delta$.

(C) The formula for the refractive index $n$ of a prism in terms of the angle of prism $A$ and the angle of minimum deviation $\delta_m$ is given by: $n = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Given that the angle of minimum deviation is equal to the angle of the prism,i.e.,$\delta_m = A$.
Substituting this into the formula: $n = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.
Using the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$,we can write $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$.
Therefore,$n = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.
Given $n = 1.5$,we have $1.5 = 2\cos(\frac{A}{2})$,which implies $\cos(\frac{A}{2}) = \frac{1.5}{2} = 0.75$.
Given $\cos(41^o) = 0.75$,we equate the angles: $\frac{A}{2} = 41^o$.
Thus,$A = 82^o$.

(A) An achromatic combination of prisms is designed to produce deviation without dispersion.
In this combination,the dispersive powers of the two prisms are chosen such that the dispersion produced by the first prism is cancelled by the second prism.
However,the deviation produced by the prisms is not cancelled.
Therefore,when white light passes through an achromatic combination,only deviation is observed.

(B) For a prism in the condition of minimum deviation,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Also,at minimum deviation,the angle of incidence $i$ is related to the prism angle $A$ and the angle of minimum deviation $\delta_m$ as $i = \frac{A + \delta_m}{2}$.
Substituting this into the refractive index formula,we get: $\mu = \frac{\sin i}{\sin(A/2)}$.
Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.
Substituting the values: $\sqrt{2} = \frac{\sin i}{\sin(60^{\circ}/2)}$.
$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.
Since $\sin 30^{\circ} = 0.5$,we have: $\sin i = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = \arcsin(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

(D) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$,where $A$ is the prism angle and $\mu$ is the refractive index of the prism material relative to the surrounding medium.
In air,$\delta_a = (_a\mu_g - 1)A$.
In water,$\delta_w = (_w\mu_g - 1)A$.
The ratio is $\frac{\delta_w}{\delta_a} = \frac{(_w\mu_g - 1)}{(_a\mu_g - 1)}$.
Given $_w\mu_g = \frac{_a\mu_g}{_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
Substituting the values: $\frac{\delta_w}{\delta_a} = \frac{(9/8 - 1)}{(3/2 - 1)} = \frac{1/8}{1/2} = \frac{1}{8} \times 2 = \frac{1}{4}$.

(A) For dispersion without deviation,the net deviation produced by the combination of two prisms must be zero.
The deviation produced by a prism is given by $\delta = A(\mu - 1)$.
For the combination of crown glass prism (angle $A$,refractive index $\mu_y$) and flint glass prism (angle $A'$,refractive index $\mu_y'$),the total deviation is $\delta_{net} = A(\mu_y - 1) + A'(\mu_y' - 1) = 0$.
Rearranging the equation to find the ratio $A'/A$:
$A'(\mu_y' - 1) = -A(\mu_y - 1)$
$\frac{A'}{A} = -\frac{(\mu_y - 1)}{(\mu_y' - 1)}$.

(C) Let the angle of the prism be $A$. When a light ray is incident by grazing one face,the angle of incidence is $90^{\circ}$,and the angle of refraction at the first surface is equal to the critical angle $C$.
In the triangle formed by the refracted ray and the two faces of the prism,the angle at the second face is $\theta$. From the geometry of the prism,$A = C + \theta$.
For the ray not to emerge from the second face,it must undergo Total Internal Reflection $(TIR)$ at that face. This requires the angle of incidence at the second face to be greater than the critical angle,i.e.,$\theta > C$.
Substituting this into the prism angle equation: $A = C + \theta > C + C$,which gives $A > 2C$.

(A) For a prism,the relation between the angle of incidence $(i_1)$,angle of emergence $(i_2)$,prism angle $(A)$,and angle of deviation $(\delta)$ is given by:
$i_1 + i_2 = A + \delta$
Given: $i_1 = 55^o$,$i_2 = 46^o$,and for an equilateral prism,$A = 60^o$.
Substituting these values:
$55^o + 46^o = 60^o + \delta$
$101^o = 60^o + \delta$
$\delta = 41^o$
We know that the angle of minimum deviation $(\delta_m)$ occurs when the angle of incidence equals the angle of emergence $(i_1 = i_2)$. Since $i_1 \neq i_2$ $(55^o \neq 46^o)$,the deviation $\delta = 41^o$ is not the minimum deviation.
Since the deviation curve is a parabola with the minimum at $i_1 = i_2$,any deviation measured at $i_1 \neq i_2$ must be greater than the minimum deviation.
Therefore,$\delta_m < \delta$,which implies $\delta_m < 41^o$.

(B) For a thin prism or at near-normal incidence,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.
Given that the angle of the prism is $\alpha$,we substitute $A = \alpha$ into the formula.
Therefore,the angle of deviation is $\delta = (\mu - 1)\alpha$.
Thus,the correct option is $B$.

(C) When a prism is placed in the position of minimum deviation,the rays of light pass through the prism symmetrically.
In this position,the refraction of light is most uniform,which minimizes the aberration effects.
As a result,the images of the spectrum formed are sharp,clear,and distinct.

(A) For a direct vision spectroscope,the net deviation produced by the combination of prisms must be zero.
Let $A_c = 9^o$ be the angle of the crown glass prisms and $A_f$ be the angle of the flint glass prisms.
The refractive indices are $\mu_c = 1.53$ and $\mu_f = 1.60$.
The total deviation $\delta_{net} = 3 \times \delta_c - 2 \times \delta_f = 0$.
Substituting the formula for deviation $\delta = (\mu - 1)A$:
$3(\mu_c - 1)A_c - 2(\mu_f - 1)A_f = 0$.
$3(1.53 - 1) \times 9^o = 2(1.60 - 1) \times A_f$.
$3(0.53) \times 9^o = 2(0.60) \times A_f$.
$14.31^o = 1.2 \times A_f$.
$A_f = \frac{14.31}{1.2} = 11.925^o \approx 11.9^o$.

(A) For dispersion without deviation,the net deviation produced by the combination must be zero.
The condition for dispersion without deviation is given by: $(\mu - 1)A + (\mu' - 1)A' = 0$.
Here,$\mu = 1.602$ (flint glass),$A = 10^{\circ}$,and $\mu' = 1.500$ (crown glass).
Taking the magnitudes,we have: $(\mu - 1)A = (\mu' - 1)A'$.
Substituting the values: $(1.602 - 1) \times 10^{\circ} = (1.500 - 1) \times A'$.
$0.602 \times 10^{\circ} = 0.500 \times A'$.
$A' = \frac{6.02}{0.500} = 12.04^{\circ}$.
Since $0.04^{\circ} = 0.04 \times 60' = 2.4'$,the angle $A' = 12^{\circ} 2.4'$.

(C) For a prism,the angle of incidence $i$ at the position of minimum deviation is given by the formula:
$i = \frac{A + \delta_m}{2}$
Given:
Angle of prism,$A = 60^o$
Angle of minimum deviation,$\delta_m = 40^o$
Substituting these values into the formula:
$i = \frac{60^o + 40^o}{2}$
$i = \frac{100^o}{2}$
$i = 50^o$
Therefore,the angle of incidence is $50^o$.

(D) In the position of minimum deviation,the light ray passes through the prism symmetrically.
According to the properties of a prism,the angle of incidence $(i)$ is equal to the angle of emergence $(e)$ when the deviation is minimum.
Therefore,$\angle i = \angle e$.

(A) hollow prism is filled with air. Since the refractive index of air is approximately $1$,the light ray passes through the prism without undergoing any significant refraction at the interfaces.
As a result,there is no change in the direction of the light (no deviation) and no splitting of white light into its constituent colors (no dispersion).
Therefore,the correct option is $A$.

(D) Given: Angle of incidence $i = 60^o$,Prism angle $A = 45^o$,Emergent angle $e = 90^o$ (relative to the surface,so the angle of emergence $e' = 0^o$ relative to the normal).
$1$. From the geometry of the prism,the angle of refraction at the second surface $r_2 = 0^o$ because the ray emerges normally.
$2$. Using the relation $A = r_1 + r_2$,we get $45^o = r_1 + 0^o$,so $r_1 = 45^o$.
$3$. Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 60^o}{\sin 45^o} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \sqrt{\frac{3}{2}}$.
$4$. The angle of deviation $\delta$ is given by $\delta = i + e' - A$. Since the ray emerges normally,the angle of emergence relative to the normal is $e' = 0^o$.
$5$. Thus,$\delta = 60^o + 0^o - 45^o = 15^o$.
Therefore,$\mu = \sqrt{\frac{3}{2}}$ and $\delta = 15^o$.

(B) For a thin 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.
Given: $\delta = 5^o$ and $\mu = 1.5$.
Substituting the values into the formula:
$5^o = (1.5 - 1)A$
$5^o = 0.5 \times A$
$A = \frac{5}{0.5} = 10^o$.
Therefore,the angle of the prism is $10^o$.

(A) The formula for the refractive index $\mu$ of a prism in terms of the angle of prism $A$ and the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$
Given $A = 60^o$ and $\delta_m = 30^o$:
$\mu = \frac{\sin((60^o + 30^o)/2)}{\sin(60^o/2)}$
$\mu = \frac{\sin(45^o)}{\sin(30^o)}$
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Therefore,the refractive index is $\sqrt{2}$.

(C) In the condition of minimum deviation,the light ray passes through the prism symmetrically.
This symmetry implies that the angle of incidence is equal to the angle of emergence.
Therefore,at the angle of minimum deviation,we have $\angle i = \angle e$ and $\angle r_1 = \angle r_2$.

(B) For dispersion without deviation,the net deviation produced by the combination of two thin prisms must be zero.
The condition for dispersion without deviation is given by: $(\mu - 1)A + (\mu' - 1)A' = 0$.
Here,$A = 4^o$,$\mu = 1.54$,and $\mu' = 1.72$.
Since the prisms are combined to produce dispersion without deviation,the deviation produced by the first prism must be balanced by the second prism in the opposite direction.
Thus,$(\mu - 1)A = -(\mu' - 1)A'$.
Taking the magnitude: $(\mu - 1)A = (\mu' - 1)A'$.
Substituting the values: $(1.54 - 1) \times 4^o = (1.72 - 1) \times A'$.
$0.54 \times 4^o = 0.72 \times A'$.
$A' = \frac{0.54 \times 4}{0.72} = \frac{2.16}{0.72} = 3^o$.
Therefore,the angle of prism $P_2$ is $3^o$.

(A) For an achromatic combination of two prisms,the net dispersion must be zero.
The condition for zero net dispersion is given by: $A(\mu_v - \mu_r) + A'(\mu'_v - \mu'_r) = 0$.
Here,$A = 10^o$,$\mu_v - \mu_r = 1.523 - 1.515 = 0.008$,and $\mu'_v - \mu'_r = 1.666 - 1.650 = 0.016$.
Substituting these values into the equation:
$10^o(0.008) + A'(0.016) = 0$.
$0.08 + A'(0.016) = 0$.
$A'(0.016) = -0.08$.
$A' = -\frac{0.08}{0.016} = -5^o$.
The magnitude of the prism angle is $5^o$ (the negative sign indicates that the second prism must be oriented in the opposite direction to the first).

(C) For a light ray to retrace its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^o$ to the surface,or $0^o$ to the normal).
In a prism,the angle of the prism $A$ is given by $A = r_1 + r_2$. Here,the ray strikes the second surface normally,so $r_2 = 0^o$.
Given $A = 30^o$,we have $30^o = r_1 + 0^o$,which implies $r_1 = 30^o$.
Applying Snell's law at the first surface:
$\mu = \frac{\sin i}{\sin r_1}$
$\sqrt{2} = \frac{\sin i}{\sin 30^o}$
$\sin i = \sqrt{2} \times \sin 30^o = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$
$i = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^o$.
Thus,the angle of incidence is $45^o$.

(A) The deviation produced by a prism 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.
Given: $i = 60^o$,$A = 30^o$,and $\delta = 30^o$.
Substituting these values into the formula: $30^o = 60^o + e - 30^o$.
Solving for $e$: $e = 30^o + 30^o - 60^o = 0^o$.
Since the angle of emergence $e$ is $0^o$,the emergent ray is normal (perpendicular) to the surface of the prism.
Therefore,the emergent ray makes an angle of $90^o$ with the face through which it emerges.

(A) For a thin prism,the angle of minimum deviation $\delta_m$ is given by the formula: $\delta_m = (\mu - 1)A$.
In a thin prism,the angle of the prism $A$ is related to the angle of refraction $r$ by the relation $A = 2r$.
Substituting the given refractive index $\mu = 1.5$ and $A = 2r$ into the formula:
$\delta_m = (1.5 - 1) \times (2r)$
$\delta_m = 0.5 \times 2r$
$\delta_m = r$.
Therefore,the correct relation is $\delta_m = r$.

(C) For a prism,the condition for minimum deviation is that the angle of incidence $(i)$ is equal to the angle of emergence $(e)$.
This symmetry implies that the refracted ray inside the prism is parallel to the base of the prism.
In the given figures,case $(3)$ shows the refracted ray inside the prism running parallel to the base of the prism.
Therefore,case $(3)$ corresponds to the condition of minimum deviation.

(D) Given that the prism is equilateral,the angle of the prism $A = 60^o$.
According to the problem,the angle of incidence $i$ is equal to the angle of emergence $e$,and both are equal to $3/4$ of the angle of the prism.
Thus,$i = e = \frac{3}{4} \times 60^o = 45^o$.
For a prism,the relationship between the angle of incidence,angle of emergence,angle of the prism,and the angle of deviation $\delta$ is given by the formula: $i + e = A + \delta$.
Substituting the known values: $45^o + 45^o = 60^o + \delta$.
$90^o = 60^o + \delta$.
Therefore,the angle of deviation $\delta = 90^o - 60^o = 30^o$.

(C) For the light ray to retrace its path,it must strike the silvered face $AC$ normally.
Let $i = 45^\circ$ be the angle of incidence at face $AB$ and $r$ be the angle of refraction.
From the geometry of the prism,the angle of the prism $A = 30^\circ$.
Since the ray strikes face $AC$ normally,the angle of incidence at face $AC$ is $0^\circ$.
In a prism,the angle $A = r_1 + r_2$. Here,$r_1 = r$ and $r_2 = 0^\circ$.
Thus,$r = A = 30^\circ$.
Using Snell's law at face $AB$: $\mu = \frac{\sin i}{\sin r} = \frac{\sin 45^\circ}{\sin 30^\circ}$.
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Therefore,the refractive index of the material of the prism is $\sqrt{2}$.

(C) For a thin prism,the angle of deviation $\delta$ is given by $\delta = (n - 1)A$,where $n$ is the refractive index and $A$ is the prism angle.
Initially,the deviation is $\delta = 34^o = (n - 1)A$.
When the shaded half is removed,the effective prism angle becomes $A' = A/2$.
The new deviation $\delta'$ will be $\delta' = (n - 1)A' = (n - 1)(A/2)$.
Substituting the initial value: $\delta' = \frac{1}{2} \times (n - 1)A = \frac{1}{2} \times 34^o = 17^o$.
Thus,the ray will suffer a deviation of $17^o$.

(B) Given: Prism angle $A = 75^o$,refractive index $\mu = \sqrt{2}$.
Let $r_1$ be the angle of refraction at the first face and $r_2$ be the angle of incidence at the second face. Since the ray is incident at the critical angle $C$ on the second face,$r_2 = C$.
We know that for a prism,$A = r_1 + r_2 = r_1 + C$.
The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$. Thus,$C = 45^o$.
Substituting the values: $75^o = r_1 + 45^o$,which gives $r_1 = 30^o$.
Applying Snell's law at the first face: $\mu = \frac{\sin i}{\sin r_1}$.
$\sqrt{2} = \frac{\sin i}{\sin 30^o} = \frac{\sin i}{0.5}$.
$\sin i = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = 45^o$.

(C) The condition for minimum deviation in a prism is that the refracted ray inside the prism must be parallel to the base of the prism.
Looking at the path of the light ray in the figure,the ray inside prism $A$ is parallel to its base,and the ray inside prism $B$ is also parallel to its base.
Therefore,the condition of minimum deviation is satisfied in prisms $A$ and $B$.

(A) The critical angle $i_c$ for the prism material is given by $\sin(i_c) = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.
Since $\sin(45^o) = \frac{1}{\sqrt{2}} \approx 0.707$,we have $\sin(45^o) > \sin(i_c)$,which implies $45^o > i_c$.
When a light ray is incident normally on the hypotenuse $BC$,it enters the prism without deviation and strikes the face $AB$ at an angle of incidence of $45^o$.
Since the angle of incidence $45^o$ is greater than the critical angle $i_c$,the ray undergoes Total Internal Reflection $(TIR)$ at face $AB$.
After reflection,the ray travels parallel to the base $BC$ and strikes the face $AC$ at an angle of incidence of $45^o$.
Again,since $45^o > i_c$,the ray undergoes $TIR$ at face $AC$ and emerges normally from the hypotenuse $BC$.
This path corresponds to the diagram shown in option $A$.

(A) The refractive index of a prism is given by the formula:
$\mu = \frac{\sin \left(\frac{A + \delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
For a very thin prism,the angle of prism $A$ and the angle of deviation $\delta$ are very small. Therefore,we can use the approximation $\sin \theta \approx \theta$.
Substituting this into the formula:
$\mu \approx \frac{\frac{A + \delta}{2}}{\frac{A}{2}}$
Simplifying the expression:
$\mu = \frac{A + \delta}{A}$
$\mu A = A + \delta$
$\delta = \mu A - A$
$\delta = (\mu - 1)A$
Thus,the angle of deviation for a thin prism is given by $\delta = (\mu - 1)A$.

(A) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$).
In the triangle formed inside the prism,the angle at the silvered surface is $90^{\circ}$,and the angle at the apex is $A$. Thus,the angle of refraction $r$ at the first surface must be $r = 90^{\circ} - A$.
According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.
Given $i = 2A$ and $r = 90^{\circ} - A$,we have:
$\mu = \frac{\sin(2A)}{\sin(90^{\circ} - A)}$
$\mu = \frac{2 \sin A \cos A}{\cos A}$
$\mu = 2 \sin A$.

(B) In the position of minimum deviation,the angle of incidence $i$ is related to the prism angle $A$ and the angle of minimum deviation $\delta_m$ by the formula:
$i = \frac{A + \delta_m}{2}$
Given:
Prism angle $A = 60^\circ$
Angle of minimum deviation $\delta_m = 30^\circ$
Substituting these values into the formula:
$i = \frac{60^\circ + 30^\circ}{2} = \frac{90^\circ}{2} = 45^\circ$
Therefore,the angle of incidence is $45^\circ$.

(A) For an equilateral prism,the angle of the prism $A = 60^{\circ}$.
In the position of minimum deviation,the angle of deviation $\delta_m$ is given as equal to the angle of the prism,so $\delta_m = A = 60^{\circ}$.
The formula for the refractive index $\mu$ of the prism material is given by:
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
Substituting the values $A = 60^{\circ}$ and $\delta_m = 60^{\circ}$:
$\mu = \frac{\sin\left(\frac{60^{\circ} + 60^{\circ}}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)}$
$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$
$\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

(D) The correct diagram is $(D)$.
When white light passes through a prism,it undergoes dispersion due to the difference in refractive indices for different wavelengths of light.
According to Cauchy's formula,the refractive index is higher for shorter wavelengths (violet) and lower for longer wavelengths (red).
Consequently,the violet light deviates the most,and the red light deviates the least.
In the correct representation,the violet ray $(V)$ should be at the bottom and the red ray $(R)$ should be at the top after emerging from the prism,as shown in diagram $(D)$.

(C) For total internal reflection,the angle of incidence $i$ must be greater than the critical angle $C$,i.e.,$i > C$.
From the figure,the ray strikes the hypotenuse at an angle of incidence $i = \theta = 45^o$.
For total internal reflection to occur,we must have $\theta > C$.
Taking the sine on both sides,$\sin \theta > \sin C$.
Since $\sin C = \frac{1}{n}$,where $n$ is the refractive index of the glass relative to air,we have $\sin \theta > \frac{1}{n}$.
Rearranging for $n$,we get $n > \frac{1}{\sin \theta}$.
Substituting $\theta = 45^o$,we get $n > \frac{1}{\sin 45^o} = \frac{1}{1/\sqrt{2}} = \sqrt{2}$.
Thus,the refractive index of the glass must be $n > \sqrt{2}$.

(A) The angular dispersion $\theta$ produced by a thin prism is given by the formula: $\theta = (\mu_b - \mu_r)A$,where $\mu_b$ is the refractive index for blue light,$\mu_r$ is the refractive index for red light,and $A$ is the prism angle.
Given:
$\mu_b = 1.66$
$\mu_r = 1.64$
$A = 5^o$
Substituting these values into the formula:
$\theta = (1.66 - 1.64) \times 5^o$
$\theta = 0.02 \times 5^o$
$\theta = 0.1^o$
Therefore,the angle of deviation between the two colours is $0.1^o$.

(B) For a combination of two thin prisms to produce dispersion without deviation,the net deviation produced by the combination must be zero.
The condition for no deviation is given by: $(\mu_1 - 1)A + (\mu_2 - 1)A' = 0$.
Here,$A = 6^o$,$\mu_1 = 1.54$,and $\mu_2 = 1.72$.
Substituting the values: $(1.54 - 1) \times 6^o + (1.72 - 1) \times A' = 0$.
$0.54 \times 6^o + 0.72 \times A' = 0$.
$3.24^o + 0.72 \times A' = 0$.
$A' = -\frac{3.24^o}{0.72} = -4.5^o$.
The negative sign indicates that the second prism is inverted relative to the first.
The magnitude of the angle is $4.5^o$,which is equal to $4^o 30'$.

(C) For an equilateral prism,the angle of the prism $A = 60^\circ$.
The formula for the refractive index $\mu$ in terms of the angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
Given $\mu = \sqrt{3}$ and $A = 60^\circ$,we substitute these values:
$\sqrt{3} = \frac{\sin\left(\frac{60^\circ + \delta_m}{2}\right)}{\sin(30^\circ)}$
Since $\sin(30^\circ) = 0.5 = \frac{1}{2}$,we have:
$\sqrt{3} \times \frac{1}{2} = \sin\left(30^\circ + \frac{\delta_m}{2}\right)$
$\frac{\sqrt{3}}{2} = \sin\left(30^\circ + \frac{\delta_m}{2}\right)$
We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$,so:
$60^\circ = 30^\circ + \frac{\delta_m}{2}$
$\frac{\delta_m}{2} = 30^\circ$
$\delta_m = 60^\circ$

(C) When the angle of incidence increases,the angle of deviation decreases until it reaches a minimum value at a specific angle of incidence.
This minimum value of the angle of deviation for a triangular prism is known as the angle of minimum deviation.
Under the condition of minimum deviation,the refracted ray inside the prism becomes parallel to the base of the prism.

(B) The formula for the refractive index of a prism is given by $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$.
Given: Prism angle $A = 60^o$,refractive index $\mu = \sqrt{2}$.
Substituting the values: $\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(60^o/2)}$.
$\sqrt{2} = \frac{\sin((60^o + \delta_m)/2)}{\sin(30^o)}$.
Since $\sin(30^o) = 0.5$,we have $\sqrt{2} \times 0.5 = \sin((60^o + \delta_m)/2)$.
$\frac{1}{\sqrt{2}} = \sin((60^o + \delta_m)/2)$.
We know that $\sin(45^o) = \frac{1}{\sqrt{2}}$,therefore $(60^o + \delta_m)/2 = 45^o$.
$60^o + \delta_m = 90^o$.
$\delta_m = 30^o$.

(B) The deviation produced by a prism is given by the formula $\delta = (\mu - 1)A$,where $\mu$ is the refractive index of the material of the prism and $A$ is the angle of the prism.
According to Cauchy's formula,the refractive index $\mu$ is inversely proportional to the square of the wavelength $\lambda$ $(\mu \propto 1/\lambda^2)$.
Therefore,light with a shorter wavelength experiences a higher refractive index and consequently suffers greater deviation.
Among the given options (Yellow,Blue,Green,Orange),Blue light has the shortest wavelength.
Thus,Blue light suffers the maximum deviation.