Refraction Through Prism Questions in English

(B) For a thin prism,the angle of minimum deviation is given by $\delta = (\mu - 1)A$.
When the prism is in air,the deviation is $\delta_a = (_a\mu_g - 1)A$.
When the prism is in water,the refractive index of glass with respect to water is $_w\mu_g = \frac{_a\mu_g}{_a\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
The deviation in water is $\delta_w = (_w\mu_g - 1)A = \left( \frac{9}{8} - 1 \right)A = \frac{1}{8}A$.
Comparing the two,$\frac{\delta_w}{\delta_a} = \frac{(1/8)A}{(1/2)A} = \frac{1}{4}$.
Therefore,the new deviation is $\delta_w = \frac{\delta_a}{4}$.

(A) The angular dispersion $\theta$ produced by a prism is given by the formula: $\theta = (\mu_v - \mu_r)A$,where $\mu_v$ is the refractive index for violet light,$\mu_r$ is the refractive index for red light,and $A$ is the angle of the prism.
Given:
$\mu_v = 1.66$
$\mu_r = 1.64$
$A = 10^o$
Substituting the values into the formula:
$\theta = (1.66 - 1.64) \times 10^o$
$\theta = 0.02 \times 10^o$
$\theta = 0.2^o$
Therefore,the angular dispersion is $0.2^o$.

(D) The refractive index $\mu$ of a prism is given by the formula:
$\mu = \frac{\sin \left( \frac{A + \delta_m}{2} \right)}{\sin \left( \frac{A}{2} \right)}$
Given:
Refracting angle $A = 60^\circ$
Angle of minimum deviation $\delta_m = 30^\circ$
Substituting these values into the formula:
$\mu = \frac{\sin \left( \frac{60^\circ + 30^\circ}{2} \right)}{\sin \left( \frac{60^\circ}{2} \right)}$
$\mu = \frac{\sin(45^\circ)}{\sin(30^\circ)}$
Since $\sin(45^\circ) = \frac{1}{\sqrt{2}}$ and $\sin(30^\circ) = \frac{1}{2}$:
$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$
$\mu \approx 1.414$

(A) The infrared spectrum of light cannot be observed using ordinary glass prisms (like flint or crown glass) because they absorb infrared radiation.
Rock salt $(NaCl)$ is transparent to infrared radiation,which allows it to transmit these wavelengths without significant absorption.
Therefore,a rock salt prism is specifically used to study the infrared spectrum of light.
Thus,the correct option is $A$.

(D) For an equilateral prism,the prism angle $A = 60^o$. When a light ray is incident normally on one surface,the angle of incidence at the first surface is $0^o$,so the angle of refraction is also $0^o$. The ray travels straight and strikes the second surface at an angle of incidence $i = A = 60^o$.
The critical angle $C$ for the material of the prism with refractive index $\mu = 1.5$ is given by:
$C = \sin^{-1}(1/\mu) = \sin^{-1}(1/1.5) = \sin^{-1}(2/3) \approx 41.8^o$.
Since the angle of incidence at the second surface $(i = 60^o)$ is greater than the critical angle $(C \approx 41.8^o)$,the condition for total internal reflection is satisfied.
Therefore,the ray undergoes total internal reflection at the second refracting surface.

(D) In the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$.
Given that the angle of incidence $i = 30^o$,therefore the angle of emergence $e = 30^o$.
The angle between the emergent ray and the second refracting surface is given by $(90^o - e)$.
Substituting the value of $e$,we get $90^o - 30^o = 60^o$.

(C) The angular dispersion $\theta$ produced by a thin prism is given by the formula: $\theta = (\mu_v - \mu_R)A$.
Given:
Angle of prism $A = 5^o$.
Refractive index for violet colour $\mu_v = 1.6$.
Refractive index for red colour $\mu_R = 1.5$.
Substituting the values in the formula:
$\theta = (1.6 - 1.5) \times 5^o$.
$\theta = 0.1 \times 5^o = 0.5^o$.
Thus,the angular dispersion produced by the prism is $0.5^o$.

(D) The angular dispersion $\theta$ produced by a thin prism is given by the formula $\theta = (\mu_v - \mu_r)A$,where $\mu_v$ and $\mu_r$ are the refractive indices for violet and red light,and $A$ is the refractive angle of the prism.
Since both prisms are made of the same material (crown glass),the term $(\mu_v - \mu_r)$ is constant for both.
Therefore,the ratio of angular dispersions is $\frac{\theta_1}{\theta_2} = \frac{A_1}{A_2}$.
Given $A_1 = 10^o$ and $A_2 = 20^o$,we have $\frac{\theta_1}{\theta_2} = \frac{10^o}{20^o} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(D) Given: Prism angle $A = 30^\circ$,refractive index $\mu = \sqrt{2}$.
Since the light ray is incident normally on one face,the angle of incidence $i = 0^\circ$,which implies the angle of refraction $r_1 = 0^\circ$.
Using the relation $A = r_1 + r_2$,we get $30^\circ = 0^\circ + r_2$,so $r_2 = 30^\circ$.
Applying Snell's law at the second surface (face $AC$): $\mu \sin r_2 = 1 \sin e$.
Substituting the values: $\sqrt{2} \sin 30^\circ = \sin e$.
$\sqrt{2} \times \frac{1}{2} = \sin e \Rightarrow \sin e = \frac{1}{\sqrt{2}}$.
Thus,the angle of emergence $e = 45^\circ$.
The angle of deviation $\delta$ is given by $\delta = e - A$ (since $i=0$ and $r_1=0$): $\delta = 45^\circ - 30^\circ = 15^\circ$.

(C) The refractive index of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$
Given that the angle of minimum deviation $\delta_m$ is equal to the angle of the prism $A$,we substitute $\delta_m = A$ into the formula:
$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$
Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:
$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$
Given $\mu = 1.732 = \sqrt{3}$,we have:
$\sqrt{3} = 2\cos(\frac{A}{2})$
$\cos(\frac{A}{2}) = \frac{\sqrt{3}}{2}$
Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$,we have $\frac{A}{2} = 30^\circ$,which implies $A = 60^\circ$.

(B) Given: Refracting angle $A = 60^\circ$,Angle of minimum deviation $\delta_m = 30^\circ$.
The refractive index $\mu$ of the prism with respect to the liquid is given by the formula:
$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$
Substituting the values:
$\mu = \frac{\sin((60^\circ + 30^\circ)/2)}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$
The relation between refractive index and critical angle $C$ is $\mu = 1/\sin(C)$.
Therefore,$\sin(C) = 1/\mu = 1/\sqrt{2}$.
This gives $C = \sin^{-1}(1/\sqrt{2}) = 45^\circ$.

(A) For a thin prism,the angle of deviation $\delta$ 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 prism angle.
Given that the prism angle $A = 60^o$ is the same for all three prisms,the angle of deviation $\delta$ depends directly on the refractive index $\mu$ such that $\delta \propto (\mu - 1)$.
For the given prisms:
$\mu_1 = 1.4 \implies \delta_1 = (1.4 - 1) \times 60^o = 0.4 \times 60^o = 24^o$
$\mu_2 = 1.5 \implies \delta_2 = (1.5 - 1) \times 60^o = 0.5 \times 60^o = 30^o$
$\mu_3 = 1.6 \implies \delta_3 = (1.6 - 1) \times 60^o = 0.6 \times 60^o = 36^o$
Comparing the values,we get $\delta_3 > \delta_2 > \delta_1$.

(C) In the position of minimum deviation for a prism,the light ray passes symmetrically through the prism.
This implies that the angle of incidence $(i_1)$ is equal to the angle of emergence $(i_2)$,i.e.,$i_1 = i_2$.
Consequently,the angle of refraction at the first surface $(r_1)$ is equal to the angle of incidence at the second surface $(r_2)$,i.e.,$r_1 = r_2$.
Since $i_1 = i_2$ and $r_1 = r_2$,the condition $i_1 = r_1$ is generally not true unless the refractive index of the prism material is $1$,which is physically impossible for a prism.
Therefore,the statement $i_1 = r_1$ is $FALSE$.

(C) The angular dispersion produced by a prism is given by $\theta = \omega \delta$,where $\omega$ is the dispersive power and $\delta$ is the mean angular deviation.
For two prisms to produce deviation without dispersion,the net angular dispersion must be zero.
$\theta_{net} = \theta + \theta' = 0$
Substituting the values,we get $\omega d + \omega' d' = 0$.
Therefore,the correct condition is $\omega d + \omega' d' = 0$.

(D) For an equilateral prism,the angle of the prism $A = 60^o$.
Given that the angle of incidence $i = 45^o$ and the angle of emergence $e = 45^o$.
The relation between the angle of incidence,angle of emergence,angle of the prism,and angle of deviation $\delta$ is given by:
$i + e = A + \delta$
Substituting the given values:
$45^o + 45^o = 60^o + \delta$
$90^o = 60^o + \delta$
$\delta = 90^o - 60^o = 30^o$.
Therefore,the angle of deviation is $30^o$.

(C) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:
$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.
Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:
$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.
Rearranging for $A$:
$\cos(\frac{A}{2}) = \frac{\mu}{2} \Rightarrow \frac{A}{2} = \cos^{-1}(\frac{\mu}{2}) \Rightarrow A = 2\cos^{-1}(\frac{\mu}{2})$.

(B) The prisms $Q$ and $R$ are of the same material and have an identical shape to prism $P$.
When these prisms are arranged as shown in the figure,the combination of prisms $Q$ and $R$ effectively acts as a glass slab with parallel faces.
$A$ light ray passing through a glass slab with parallel faces undergoes lateral displacement but no net angular deviation.
Therefore,the total deviation experienced by the light ray remains the same as the deviation caused by prism $P$ alone,which is the minimum deviation.

(C) The angle of a prism is defined as the angle between the two refracting surfaces of the prism.
In the given figure,the light enters through one face and emerges from the other. The two refracting surfaces meet at the vertex $C$.
Therefore,the angle at vertex $C$ represents the angle of the prism.

(A) For a prism,the condition for minimum deviation is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
At minimum deviation,the angle of incidence $i$ is related to the refractive index $\mu$ and the prism angle $A$ by the relation: $\mu = \frac{\sin i}{\sin(A/2)}$.
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,$i = \arcsin(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

(A) In the condition of minimum deviation,the angle of refraction $r$ is related to the angle of the prism $A$ by the formula $r = \frac{A}{2}$.
Given that the angle of the prism $A = 60^o$.
Substituting the value,we get $r = \frac{60^o}{2} = 30^o$.
Therefore,the angle of refraction is $30^o$.

(B) In the position of minimum deviation,the refracted ray inside the prism is parallel to the base of the prism.
Since the prism is placed on a horizontal table,its base is horizontal.
Therefore,for the refracted ray $QR$ to be parallel to the base,$QR$ must also be horizontal.

(A) The deviation $\delta$ produced by a prism is given by the relation $\delta = (\mu - 1)A$, 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 $\lambda$ $(\mu \propto \frac{1}{\lambda^2})$.
Since the wavelength of red light is the longest among the visible spectrum, it experiences the least refractive index $\mu$.
Consequently, red light undergoes the minimum deviation $\delta$ compared to other colours.

(A) The formula for the refractive index of a prism in the case of minimum deviation is given by: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$
Given: Refracting angle $A = 60^{\circ}$ and minimum deviation $\delta_m = 38^{\circ}$.
Substituting these values into the formula:
$\mu = \frac{\sin(\frac{60^{\circ} + 38^{\circ}}{2})}{\sin(\frac{60^{\circ}}{2})}$
$\mu = \frac{\sin(\frac{98^{\circ}}{2})}{\sin(30^{\circ})}$
$\mu = \frac{\sin(49^{\circ})}{\sin(30^{\circ})}$
Using the value $\sin(49^{\circ}) \approx 0.7547$ and $\sin(30^{\circ}) = 0.5$:
$\mu = \frac{0.7547}{0.5} = 1.5094 \approx 1.5$.

(D) The formula for the deviation produced by a prism is given by $\delta = i_1 + i_2 - A$,where $\delta$ is the angle of deviation,$i_1$ is the angle of incidence,$i_2$ is the angle of emergence,and $A$ is the angle of the prism.
Given values are $\delta = 55^o$,$i_1 = 15^o$,and $A = 60^o$.
Substituting these values into the formula: $55^o = 15^o + i_2 - 60^o$.
$55^o = i_2 - 45^o$.
$i_2 = 55^o + 45^o = 100^o$.
Since the angle of emergence $i_2$ must be less than the critical angle for the prism material to emerge,an angle of $100^o$ is physically impossible for a standard prism refraction scenario. Therefore,the correct option is $D$.

(D) For an achromatic combination,the net dispersion is zero,which implies the dispersive powers satisfy the condition: $(\mu_{v,C} - \mu_{r,C})A_C = -(\mu_{v,F} - \mu_{r,F})A_F$.
This rearranges to: $\mu_{r,C}A_C + \mu_{r,F}A_F = \mu_{v,C}A_C + \mu_{v,F}A_F$.
Substituting the given values:
$\mu_{r,C}A_C + \mu_{r,F}A_F = (1.5 \times 19) + (1.66 \times 6) = 28.5 + 9.96 = 38.46 \approx 38.5$.
The resultant deviation for red light is given by: $\delta_r = (\mu_{r,C} - 1)A_C + (\mu_{r,F} - 1)A_F$.
$\delta_r = (\mu_{r,C}A_C + \mu_{r,F}A_F) - (A_C + A_F)$.
$\delta_r = 38.5 - (19 + 6) = 38.5 - 25 = 13.5^o$.

(D) The prism is isosceles with an apex angle of $120^\circ$. The base angles are $(180^\circ - 120^\circ)/2 = 30^\circ$.
Let the incident rays be normal to the first face. However,based on the geometry,the rays enter the prism and refract. Let the angle of incidence at the first surface be $i = 30^\circ$ relative to the normal.
Applying Snell's law at the first surface: $1 \cdot \sin(30^\circ) = 1.44 \cdot \sin(r)$.
$\sin(r) = \frac{0.5}{1.44} = \frac{50}{144} = \frac{25}{72} \approx 0.347$.
Wait,looking at the provided solution geometry: The angle of incidence at the prism face is $30^\circ$ relative to the surface normal. Thus,$\sin(30^\circ) = 1.44 \sin(r) \Rightarrow \sin(r) = 0.5 / 1.44 = 0.347$.
Actually,the provided solution uses $\sin(r) = 0.72$. This implies the angle of incidence was $i$ such that $\sin(i) = 1.44 \times 0.72 = 1.036$,which is impossible.
Re-evaluating the geometry: The angle of the prism is $120^\circ$. The rays are incident normally on the first face. The angle of refraction $r$ at the second face is $30^\circ$. Using Snell's law: $1.44 \sin(30^\circ) = 1 \sin(e) \Rightarrow \sin(e) = 1.44 \times 0.5 = 0.72$.
Thus,$e = \sin^{-1}(0.72)$. The deviation of each ray is $\delta = e - r = \sin^{-1}(0.72) - 30^\circ$.
The total angle between the two emerging rays is $2\delta = 2\{\sin^{-1}(0.72) - 30^\circ\}$.

(D) Let the angle of incidence at the first surface be $i$. Since the ray passes through the prisms without deviation,the angle of emergence from the last surface must also be $i$.
Applying Snell's Law at each interface:
$1$. At the first interface (air to prism $n_1$): $1 \cdot \sin i = n_1 \cdot \sin r_1 \implies \sin^2 i = n_1^2 \sin^2 r_1$ ... $(i)$
$2$. At the interface between prism $n_1$ and $n_2$: $n_1 \cdot \sin(90^\circ - r_1) = n_2 \cdot \sin r_2 \implies n_1 \cos r_1 = n_2 \sin r_2 \implies n_1^2 \cos^2 r_1 = n_2^2 \sin^2 r_2$ ... $(ii)$
$3$. At the interface between prism $n_2$ and $n_3$: $n_2 \cdot \sin(90^\circ - r_2) = n_3 \cdot \sin r_3 \implies n_2 \cos r_2 = n_3 \sin r_3 \implies n_2^2 \cos^2 r_2 = n_3^2 \sin^2 r_3$ ... $(iii)$
$4$. At the last interface (prism $n_3$ to air): $n_3 \cdot \sin(90^\circ - r_3) = 1 \cdot \sin i \implies n_3 \cos r_3 = \sin i \implies n_3^2 \cos^2 r_3 = \sin^2 i$ ... $(iv)$
Adding $(i)$,$(ii)$,$(iii)$,and $(iv)$:
$\sin^2 i + n_1^2 \cos^2 r_1 + n_2^2 \cos^2 r_2 + n_3^2 \cos^2 r_3 = n_1^2 \sin^2 r_1 + n_2^2 \sin^2 r_2 + n_3^2 \sin^2 r_3 + \sin^2 i$
Using $\sin^2 \theta + \cos^2 \theta = 1$,this simplifies to:
$n_1^2 + n_3^2 = 1 + n_2^2$.

(A) For a prism,the relationship between the angle of deviation $(\delta)$ and the angle of incidence $(i)$ is given by the formula $\delta = (i + e) - A$,where $e$ is the angle of emergence and $A$ is the angle of the prism.
As the angle of incidence $(i)$ increases,the angle of deviation $(\delta)$ initially decreases.
It reaches a minimum value known as the angle of minimum deviation $(\delta_m)$ at a specific angle of incidence.
After this point,as the angle of incidence $(i)$ continues to increase,the angle of deviation $(\delta)$ starts to increase again.
Therefore,the graph between $\delta$ and $i$ is a parabolic-like curve that shows a minimum,which corresponds to the graph shown in option $A$.

(D) For a small angled prism,the angle of minimum deviation $\delta_m$ is given by the formula:
$\delta_m = (\mu - 1)A$
This equation represents a straight line of the form $y = mx + c$,where $y = \delta_m$,$x = \mu$,$m = A$,and $c = -A$.
At point $P$,the deviation $\delta_m = 0$. Substituting this into the formula:
$0 = (\mu - 1)A$
Since $A \neq 0$,we have $\mu - 1 = 0$,which implies $\mu = 1$. Thus,point $P$ corresponds to $\mu = 1$.
Comparing $\delta_m = A\mu - A$ with the linear equation $y = mx + c$,the slope $m$ is equal to $A$.
Therefore,both statements $(a)$ and $(c)$ are correct.

(C) Given: Prism angle $A = 5^o$,refractive index for red light $\mu_R = 1.5$,and refractive index for violet light $\mu_V = 1.6$.
The formula for angular dispersion $\theta$ produced by a thin prism is given by:
$\theta = (\mu_V - \mu_R) A$
Substituting the given values:
$\theta = (1.6 - 1.5) \times 5^o$
$\theta = 0.1 \times 5^o$
$\theta = 0.5^o$
Therefore,the angular dispersion produced by the prism is $0.5^o$.

(C) Given: Refractive index of glass with respect to air $^{a}\mu_{g} = 1.53$,refractive index of water with respect to air $^{a}\mu_{w} = 1.33$,and prism angle $A = 60^{\circ}$.
The refractive index of the prism with respect to water is given by:
$^{w}\mu_{g} = \frac{^{a}\mu_{g}}{^{a}\mu_{w}} = \frac{1.53}{1.33} \approx 1.15$.
The formula for the angle of minimum deviation $\delta_{m}$ is:
$^{w}\mu_{g} = \frac{\sin(\frac{A + \delta_{m}}{2})}{\sin(\frac{A}{2})}$.
Substituting the values:
$1.15 = \frac{\sin(\frac{60^{\circ} + \delta_{m}}{2})}{\sin(30^{\circ})}$.
Since $\sin(30^{\circ}) = 0.5$:
$\sin(\frac{60^{\circ} + \delta_{m}}{2}) = 1.15 \times 0.5 = 0.575$.
Taking the inverse sine:
$\frac{60^{\circ} + \delta_{m}}{2} = \sin^{-1}(0.575) \approx 35.1^{\circ}$.
Solving for $\delta_{m}$:
$60^{\circ} + \delta_{m} = 70.2^{\circ} \implies \delta_{m} = 10.2^{\circ}$.

(B) For a light ray passing through a prism,the condition for minimum deviation is that the angle of incidence is equal to the angle of emergence $(i = e)$.
In this symmetric condition,the refracted ray inside the prism $(QR)$ becomes parallel to the base of the prism.
Since the prism is placed on a horizontal table,its base is horizontal.
Therefore,for minimum deviation,the refracted ray $QR$ inside the prism must be horizontal.

(A) For a thin prism with a small refracting angle $A$ and refractive index $\mu$,the angle of deviation $\delta$ is given by the formula:
$\delta = (\mu - 1) A$
This formula is derived from the general prism formula $\delta = i + e - A$. For small angles,the angles of incidence and refraction are small,leading to the approximation $\delta = (\mu - 1) A$.

(C) Given,refractive index $\mu = \sqrt{2}$ and prism angle $A = 30^\circ$.
Since the light ray retraces its path after reflection from the polished surface,it must strike the polished surface normally.
Therefore,the angle of refraction at the second surface is $r_2 = 0^\circ$.
From the relation $A = r_1 + r_2$,we have $30^\circ = r_1 + 0^\circ$,which gives $r_1 = 30^\circ$.
Applying Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$.
$\sqrt{2} = \frac{\sin i}{\sin 30^\circ}$.
$\sin i = \sqrt{2} \times \sin 30^\circ = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Thus,$i = 45^\circ$.

(C) For a triangular prism,the angle of deviation $(\delta)$ varies with the angle of incidence $(i)$.
As the angle of incidence $(i)$ increases,the angle of deviation $(\delta)$ first decreases,reaches a minimum value known as the angle of minimum deviation $(\delta_m)$,and then increases.
This relationship is represented by a $U$-shaped curve,which is characteristic of the deviation versus incidence angle graph for a prism.

(A) For a prism,the angle of deviation $(\delta)$ depends on the angle of incidence $(i)$.
As the angle of incidence $(i)$ increases,the angle of deviation $(\delta)$ first decreases,reaches a minimum value (known as the angle of minimum deviation,$\delta_m$),and then starts increasing.
This relationship is represented by a parabolic-like curve where $\delta$ is plotted on the $y$-axis and $i$ is plotted on the $x$-axis.
The graph starts at a certain value of $\delta$,decreases to a minimum,and then increases again.
This corresponds to the curve shown in the first option (image $179-$a126).

(A) In a prism,the deviation angle $\delta$ is given by $\delta = (i_1 + i_2) - A$,where $A$ is the prism angle.
For minimum deviation,the condition is that the angle of incidence must be equal to the angle of emergence.
Therefore,$i_1 = i_2$.
At this condition,the light ray inside the prism passes parallel to the base of the prism.

(C) For dispersion without deviation,the condition is $(\mu - 1)A + (\mu' - 1)A' = 0$.
Given: $A = 15^\circ$,$\mu = 1.65$,$\omega = 0.03$ (flint) and $\mu' = 1.52$,$\omega' = 0.02$ (crown).
Substituting the values: $(1.65 - 1) \times 15^\circ + (1.52 - 1) \times A' = 0$.
$0.65 \times 15^\circ + 0.52 \times A' = 0$.
$A' = -\frac{0.65 \times 15^\circ}{0.52} = -18.75^\circ$.
The negative sign indicates that the prisms are placed in opposition.
The net angular dispersion is given by $\delta_{\theta} = \omega(\mu - 1)A + \omega'(\mu' - 1)A'$.
Since $(\mu - 1)A = -(\mu' - 1)A'$,we have $\delta_{\theta} = \omega(\mu - 1)A - \omega'(\mu - 1)A = (\omega - \omega')(\mu - 1)A$.
$\delta_{\theta} = (0.03 - 0.02) \times (1.65 - 1) \times 15^\circ$.
$\delta_{\theta} = 0.01 \times 0.65 \times 15^\circ = 0.0975^\circ$.

(C) Given: Angle of incidence $i = 60^{\circ}$,Angle of prism $A = 30^{\circ}$,Angle of deviation $\delta = 30^{\circ}$.
Using the formula for the angle of deviation: $\delta = i + e - A$.
Substituting the values: $30^{\circ} = 60^{\circ} + e - 30^{\circ}$.
Solving for the angle of emergence: $e = 0^{\circ}$.
Since $e = 0^{\circ}$,the emergent ray is normal to the second surface,so the angle of refraction at the second surface $r_2 = 0^{\circ}$.
Using the relation $A = r_1 + r_2$: $30^{\circ} = r_1 + 0^{\circ}$,which gives $r_1 = 30^{\circ}$.
Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1}$.
$\mu = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(A) For an equilateral prism,the prism angle $A = 60^\circ$.
Given that the angle of minimum deviation $\delta_m = A = 60^\circ$.
At the condition of minimum deviation,the angle of incidence $i$ is given by the formula:
$i = \frac{A + \delta_m}{2}$
Substituting the values:
$i = \frac{60^\circ + 60^\circ}{2} = \frac{120^\circ}{2} = 60^\circ$.
Therefore,the angle of incidence is $60^\circ$.

(B) For a prism,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:
$A = 60^{\circ}$
$\delta_m = 30^{\circ}$
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) Given:
Angle of prism,$A = 6^{\circ}$
Refractive index,$\mu = 1.5$
For a thin prism,the angle of minimum deviation $\delta_m$ is given by the formula:
$\delta_m = (\mu - 1)A$
Substituting the given values:
$\delta_m = (1.5 - 1) \times 6^{\circ}$
$\delta_m = 0.5 \times 6^{\circ}$
$\delta_m = 3^{\circ}$
Therefore,the angle of minimum deviation is $3^{\circ}$.

(A) In a prism,the angle of deviation $\delta$ is given by the relation $\delta = i + e - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is the prism angle.
At the condition of minimum deviation,the light ray passes symmetrically through the prism.
This symmetry implies that the angle of incidence is equal to the angle of emergence,i.e.,$i = e$.
Therefore,at minimum deviation,the angle of emergence is equal to the angle of incidence.

(C) For a prism,the condition for minimum deviation is that 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 $A = 40^{\circ}$ and angle of incidence $i = 38^{\circ}$.
Substituting the values into the formula:
$38 = \frac{40 + \delta_m}{2}$
$76 = 40 + \delta_m$
$\delta_m = 76 - 40 = 36^{\circ}$.
Thus,the angle of minimum deviation is $36^{\circ}$.

(D) Since the incident ray retraces its path,it must strike the silvered face of the prism normally (at an angle of $90^\circ$).
In the triangle formed by the prism,let the vertices be $A, B, C$ and the point of incidence be $D$ on face $AC$,and the point on the silvered face be $E$. In $\triangle AED$,the angles are $30^\circ$,$90^\circ$,and $\angle ADE$.
Sum of angles in $\triangle AED = 180^\circ$.
$30^\circ + 90^\circ + \angle ADE = 180^\circ \Rightarrow \angle ADE = 60^\circ$.
The angle of refraction $r$ is the angle between the normal at $D$ and the refracted ray $DE$. Since the normal is perpendicular to the face $AC$,we have $\angle r + \angle ADE = 90^\circ$.
$\angle r = 90^\circ - 60^\circ = 30^\circ$.
Applying Snell's Law at the surface $AC$:
$1 \cdot \sin i = \mu \cdot \sin r$
$\sin i = \sqrt{2} \cdot \sin 30^\circ$
$\sin i = \sqrt{2} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = 45^\circ$.

(A) Given: Prism angle $A = 60^\circ$,Refractive index $\mu = \sqrt{2}$.
The formula for the refractive index of a prism in terms of the angle of minimum deviation $\delta_m$ is:
$\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$
Substituting the values:
$\sqrt{2} = \frac{\sin((60^\circ + \delta_m)/2)}{\sin(60^\circ/2)}$
$\sqrt{2} = \frac{\sin(30^\circ + \delta_m/2)}{\sin(30^\circ)}$
Since $\sin(30^\circ) = 0.5$:
$\sqrt{2} \times 0.5 = \sin(30^\circ + \delta_m/2)$
$\frac{1}{\sqrt{2}} = \sin(30^\circ + \delta_m/2)$
We know that $\sin(45^\circ) = \frac{1}{\sqrt{2}}$,therefore:
$45^\circ = 30^\circ + \delta_m/2$
$15^\circ = \delta_m/2$
$\delta_m = 30^\circ$.

(C) The formula for the refractive index of a prism in the condition of minimum deviation is given by: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.
Substituting the values: $\sqrt{2} = \frac{\sin(\frac{60^{\circ} + \delta_m}{2})}{\sin(30^{\circ})}$.
Since $\sin(30^{\circ}) = 0.5$,we have: $\sqrt{2} \times 0.5 = \sin(\frac{60^{\circ} + \delta_m}{2})$.
$\frac{\sqrt{2}}{2} = \sin(\frac{60^{\circ} + \delta_m}{2}) \Rightarrow \sin(45^{\circ}) = \sin(\frac{60^{\circ} + \delta_m}{2})$.
Equating the angles: $45^{\circ} = \frac{60^{\circ} + \delta_m}{2} \Rightarrow 90^{\circ} = 60^{\circ} + \delta_m \Rightarrow \delta_m = 30^{\circ}$.
In the condition of minimum deviation,the angle of incidence $i$ is given by: $i = \frac{A + \delta_m}{2}$.
$i = \frac{60^{\circ} + 30^{\circ}}{2} = \frac{90^{\circ}}{2} = 45^{\circ}$.

(D) For an equilateral prism,the angle of the prism $A = 60^o$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and $i = e = (3/4)A$.
Substituting the value of $A$:
$i = e = (3/4) \times 60^o = 45^o$.
The formula for the angle of deviation $\delta$ is given by:
$\delta = i + e - A$.
Substituting the known values:
$\delta = 45^o + 45^o - 60^o$.
$\delta = 90^o - 60^o = 30^o$.
Thus,the angle of deviation is $30^o$.