Ray Optics and Optical Instruments

Using the lens maker's formula

$\frac{1}{f_1}=\left(n_{\text{glass }}-n_{\text{air }}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Since one surface is flat, $R_2$ is infinity, so

$\frac{1}{R_2}=0$

For the first lens in air (with refractive index 1.5 and radius of curvature 2 cm ):

$$\begin{gathered} \frac{1}{f_1}=(1.5-1)\left(\frac{1}{2}\right) \\ {f}_1=4cm \end{gathered}$$

For the second lens:

$$\begin{gathered} \frac{1}{f_2}=\left(\frac{1.5}{1.2}-1\right)\left(\frac{1}{3}-0\right) \\ f_2=12cm \end{gathered}$$

Ratio of focal lengths:

$f_1:f_2=4:12=1:3$

To determine the nature of the image formed by a convex spherical surface separating air and glass, we need to consider the refraction of light at the convex surface. The nature of the image (real or virtual) depends on the position of the object relative to the focal point of the convex surface.

Identify the given parameters: object distance (\(u = -4R\)), radius of curvature (\(R\)), and the refractive indices of air (\(n1 = 1\)) and glass (\(n2 > 1\)).

Use the refraction formula for a spherical surface:
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\)
where \(u\) is the object distance, \(v\) is the image distance, and \(R\) is the radius of curvature.

Substitute the given values into the formula:
\(\frac{n_2}{v} - \frac{1}{-4R} = \frac{n_2 - 1}{R}\)

Solve for the image distance \(v\). As the object moves towards the surface (\(u\) increases from \(-4R\) to \(0\)), analyze how the image distance \(v\) changes. Initially, when the object is far (\(u = -4R\)), the image is virtual. As the object moves closer to the surface, the image becomes real when the object is within the focal length of the convex surface.

Final Answer:
D. first real and then virtual

The apparent depth ($d_a$) is given by the formula:

$d_a=\frac{d}{n}$​

where $d$is the actual depth of the coin and $n$ is the refractive index of the medium (water in this case).

$d_a=\frac{H}{\frac{4}{3}}$​

To solve this, multiply $H$ by the reciprocal of $\frac{4}{3}$:

$d_a=H\times \frac{3}{4}=\frac{3H}{4}$​

There is no effect of medium on focal length of Spherical mirror.

The apparent depth ($d_a$) is given by the formula:

$d_a=\frac{d}{n}$​

where $d$is the actual depth of the coin and $n$ is the refractive index of the medium (water in this case).

$d_a=\frac{H}{\frac{4}{3}}$​

To solve this, multiply $H$ by the reciprocal of $\frac{4}{3}$:

$d_a=H\times \frac{3}{4}=\frac{3H}{4}$​

(a) The combined focal length of a lens and a mirror can be derived as:

$\frac{1}{f_{eq}}=\frac{2}{f_L}-\frac{1}{f_m}$

The focal length of the mirror is given as:

$f_m=\frac{-\left|R_2\right|}{2}$

Thus, the equivalent focal length becomes:

$\frac{1}{f_{eq}}=\frac{2}{f_L}+\frac{2}{R_2}$

The focal length $f_L$ of the lens:

$\frac{1}{f_L}=\left(\frac{{\mu }_2-{\mu }_1}{{\mu }_1}\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

Combining the expressions, we obtain the simplified formula for the equivalent focal length:

$\frac{1}{f_{eq}}=\frac{2\left({\mu }_2{R}_1+{\mu }_2{R}_2-{\mu }_1{R}_2\right)}{{\mu }_1{R}_1{R}_2}$

For same size of image

$u=2f$

Substituting the focal length formula:

$u=\frac{{\mu }_1{R}_1{R}_2}{{\mu }_2{R}_1+{\mu }_2{R}_2-{\mu }_1{R}_2}$

$$\begin{gathered} PO=u=-x \\ PI=v=x \\ PO=PI \end{gathered}$$

Using the refraction formula for a spherical surface separating two media, we have:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Substituting the values for the refractive indices:

$\frac{1.5}{x}+\frac{1}{-x}=\frac{1.5-1}{R}$

Simplifying:

$$\begin{gathered} \frac{5}{2x}=\frac{1}{2R} \\ x=5R \end{gathered}$$

The focal length F is the reciprocal of optical power P :

$F=\frac{1}{P}$

When P =2.5 D,

$F=\frac{1}{2.5}=\frac{2}{5}$

When P increases to 2.6 D ,

$F^{\text{'}}=\frac{1}{2.6}=\frac{5}{13}$

Relative decrease in focal length:

$\frac{F-F^{\text{'}}}{F}=\frac{\frac{2}{5}-\frac{5}{13}}{\frac{2}{5}}$

Simplifying:

$=\frac{26-25}{26}=\frac{1}{26}\approx 0.04$

the correct answer is 0.04 .

There is no effect of medium on focal length of Spherical mirror.

Using the lens maker's formula

$\frac{1}{f_1}=\left(n_{\text{glass }}-n_{\text{air }}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Since one surface is flat, $R_2$ is infinity, so

$\frac{1}{R_2}=0$

For the first lens in air (with refractive index 1.5 and radius of curvature 2 cm ):

$$\begin{gathered} \frac{1}{f_1}=(1.5-1)\left(\frac{1}{2}\right) \\ {f}_1=4cm \end{gathered}$$

For the second lens:

$$\begin{gathered} \frac{1}{f_2}=\left(\frac{1.5}{1.2}-1\right)\left(\frac{1}{3}-0\right) \\ f_2=12cm \end{gathered}$$

Ratio of focal lengths:

$f_1:f_2=4:12=1:3$

To determine the conditions required for total internal reflection at the interface between two media where the speed of light in medium ' 1 ' is \(v_1\) and in medium ' 2 ' is \(v_2\left(v_2>v_1\right)\), consider Snell's Law:

\(n_1 \sin \theta_1=n_2 \sin \theta_2\)

where \(n_1\) and \(n_2\) are the refractive indices of media ' 1 ' and ' 2 '. The refractive index is inversely proportional to the speed of light in the medium:

\(n=\frac{c}{v}\)

where \(c\) is the speed of light in a vacuum. Thus, \(n_1=\frac{c}{v_1}\) and \(n_2=\frac{c}{v_2}\). Total internal reflection occurs when light is incident in a denser medium and attempts to pass into a lighter medium (from medium ' 1 ' to medium ' 2 ', \(v_2>v_1\), implying \(n_2<n_1\) ). For total internal reflection, \(\theta_2\) must reach \(90^{\circ}\), thus:

\(n_1 \sin \theta_c=n_2 \sin 90^{\circ}\)

where \(\theta_c\) is the critical angle, leading to:

\(\sin \theta_c=\frac{n_2}{n_1}\)

Since refractive indices are inversely proportional to velocity:

\(\sin \theta_c=\frac{v_1}{v_2}\)

To achieve total internal reflection, the incident angle \(\theta_1\) must be greater than this critical angle:

\(\theta_1>\sin ^{-1}\left(\frac{v_1}{v_2}\right)\)

Therefore, for total internal reflection, the ray must be incident from medium ' 1 ' at an angle greater than \(\sin ^{-1}\left(\frac{v_1}{v_2}\right)\).

According to the formula of silvering of lens

$$\begin{gathered} -\frac{1}{\mathrm{F}}=\frac{2}{f_L}-\frac{1}{f_m} \\ -\frac{1}{\mathrm{F}}=2\left(\frac{{\mathrm{\mu }}_2}{{\mathrm{\mu }}_1}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{-{\mathrm{R}}_2}\right)-\left(\frac{1}{-{\displaystyle \frac{{\mathrm{R}}_2}{2}}}\right) \\ -\frac{1}{\mathrm{F}}=2\left(\frac{{\mathrm{\mu }}_2-{\mathrm{\mu }}_1}{{\mathrm{\mu }}_1}\right)\left(\frac{{\mathrm{R}}_1 + {\mathrm{R}}_2}{{\mathrm{R}}_1 {\mathrm{R}}_2}\right)+\left(\frac{2}{{\mathrm{R}}_2}\right)\left(\frac{{\mathrm{\mu }}_1 {\mathrm{R}}_1}{{\mathrm{\mu }}_1 {\mathrm{R}}_1}\right) \\ -\frac{1}{\mathrm{F}}=\frac{2}{{\mathrm{\mu }}_1 {\mathrm{R}}_1{\mathrm{R}}_2}\left[\left({\mathrm{\mu }}_2-{\mathrm{\mu }}_1\right)\left({\mathrm{R}}_1 + {\mathrm{R}}_2\right)+{\mathrm{\mu }}_1 {\mathrm{R}}_1\right] \\ -\frac{1}{\mathrm{F}}=\frac{2}{{\mathrm{\mu }}_1 {\mathrm{R}}_1{\mathrm{R}}_2}\left[{\mathrm{\mu }}_2\left({\mathrm{R}}_1 + {\mathrm{R}}_2\right)-{\mathrm{\mu }}_1 {\mathrm{R}}_2\right] \\ \left|-2\mathrm{F}\right|=\frac{{\mathrm{\mu }}_1 {\mathrm{R}}_1{\mathrm{R}}_2}{\left[{\mathrm{\mu }}_2\left({\mathrm{R}}_1 + {\mathrm{R}}_2\right)-{\mathrm{\mu }}_1 {\mathrm{R}}_2\right]} \end{gathered}$$

• Focal length $f$ is related to power $P$ by $f = \frac{100}{P}$.
• Initial focal length $f_1 = \frac{100}{0.1} = 1000 \, \text{cm}$
• Final focal length $f_2 = \frac{100}{2.5} = 40 \, \text{cm}$
• Relative shift: $\frac{f_1-f_2}{f_1}=\frac{960}{1000}=\frac{24}{25}\mathrm{.}$

Thus, the relative shift in the focal length is $\frac{24}{25}$​.

We are given:

• Distance between object and image = 20 cm
• Magnification ($m$) = -3
• We need to find the focal length (f).

The magnification formula for mirrors is:

$$m = \frac{-v}{u}$$​

Given $m = -3$, we get:

$$-3 = \frac{-v}{u}$$

$$v = 3u$$

Also, the total distance between the object and image is:

$$|v - u| = 20$$

Substituting $v = 3u$:

$$3u - u = 20$$

$$2u = 20$$

$$u = 10 \text{ cm}$$

$$v = 3(10) = 30 \text{ cm}$$

The mirror equation is:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$​

Substituting $u = 10$ and $v = 30$

$$\frac{1}{f} = \frac{1}{10} + \frac{1}{30}$$

​ $$\frac{1}{f} = \frac{3+1}{30} = \frac{4}{30} = \frac{2}{15}$$

​ $$f = \frac{15}{2} = 7.5 \text{ cm}$$

$$\boxed{f = 7.5 \text{ cm}}$$

To determine the conditions required for total internal reflection at the interface between two media where the speed of light in medium ' 1 ' is \(v_1\) and in medium ' 2 ' is \(v_2\left(v_2>v_1\right)\), consider Snell's Law:

\(n_1 \sin \theta_1=n_2 \sin \theta_2\)

where \(n_1\) and \(n_2\) are the refractive indices of media ' 1 ' and ' 2 '. The refractive index is inversely proportional to the speed of light in the medium:

\(n=\frac{c}{v}\)

where \(c\) is the speed of light in a vacuum. Thus, \(n_1=\frac{c}{v_1}\) and \(n_2=\frac{c}{v_2}\). Total internal reflection occurs when light is incident in a denser medium and attempts to pass into a lighter medium (from medium ' 1 ' to medium ' 2 ', \(v_2>v_1\), implying \(n_2<n_1\) ). For total internal reflection, \(\theta_2\) must reach \(90^{\circ}\), thus:

\(n_1 \sin \theta_c=n_2 \sin 90^{\circ}\)

where \(\theta_c\) is the critical angle, leading to:

\(\sin \theta_c=\frac{n_2}{n_1}\)

Since refractive indices are inversely proportional to velocity:

\(\sin \theta_c=\frac{v_1}{v_2}\)

To achieve total internal reflection, the incident angle \(\theta_1\) must be greater than this critical angle:

\(\theta_1>\sin ^{-1}\left(\frac{v_1}{v_2}\right)\)

Therefore, for total internal reflection, the ray must be incident from medium ' 1 ' at an angle greater than \(\sin ^{-1}\left(\frac{v_1}{v_2}\right)\).

To determine the conditions required for total internal reflection at the interface between two media where the speed of light in medium ' 1 ' is \(v_1\) and in medium ' 2 ' is \(v_2\left(v_2>v_1\right)\), consider Snell's Law:

\(n_1 \sin \theta_1=n_2 \sin \theta_2\)

where \(n_1\) and \(n_2\) are the refractive indices of media ' 1 ' and ' 2 '. The refractive index is inversely proportional to the speed of light in the medium:

\(n=\frac{c}{v}\)

where \(c\) is the speed of light in a vacuum. Thus, \(n_1=\frac{c}{v_1}\) and \(n_2=\frac{c}{v_2}\). Total internal reflection occurs when light is incident in a denser medium and attempts to pass into a lighter medium (from medium ' 1 ' to medium ' 2 ', \(v_2>v_1\), implying \(n_2<n_1\) ). For total internal reflection, \(\theta_2\) must reach \(90^{\circ}\), thus:

\(n_1 \sin \theta_c=n_2 \sin 90^{\circ}\)

where \(\theta_c\) is the critical angle, leading to:

\(\sin \theta_c=\frac{n_2}{n_1}\)

Since refractive indices are inversely proportional to velocity:

\(\sin \theta_c=\frac{v_1}{v_2}\)

To achieve total internal reflection, the incident angle \(\theta_1\) must be greater than this critical angle:

\(\theta_1>\sin ^{-1}\left(\frac{v_1}{v_2}\right)\)

Therefore, for total internal reflection, the ray must be incident from medium ' 1 ' at an angle greater than \(\sin ^{-1}\left(\frac{v_1}{v_2}\right)\).

$$\begin{gathered} PO=u=-x \\ PI=v=x \\ PO=PI \end{gathered}$$

Using the refraction formula for a spherical surface separating two media, we have:

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

Substituting the values for the refractive indices:

$\frac{1.5}{x}+\frac{1}{-x}=\frac{1.5-1}{R}$

Simplifying:

$$\begin{gathered} \frac{5}{2x}=\frac{1}{2R} \\ x=5R \end{gathered}$$

The apparent depth ($d_a$) is given by the formula:

$d_a=\frac{d}{n}$​

where $d$is the actual depth of the coin and $n$ is the refractive index of the medium (water in this case).

$d_a=\frac{H}{\frac{4}{3}}$​

To solve this, multiply $H$ by the reciprocal of $\frac{4}{3}$:

$d_a=H\times \frac{3}{4}=\frac{3H}{4}$​

$\mu =\frac{\sin \left({\displaystyle \raisebox{1ex}{$A+{\delta }_m$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\sin \left({\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}$

$\frac{\cos \left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{\sin \left(\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}=\frac{\sin \left({\displaystyle \raisebox{1ex}{$A+\delta m$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\sin \left({\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}$

${\delta }_m=\mathrm{\pi }-2\mathrm{A}$

The formula for the lateral shift x of a ray passing through a parallel-sided slab is given by:

$x=\frac{h\sin (i-r)}{\cos \left(r\right)}$

The focal length F is the reciprocal of optical power P :

$F=\frac{1}{P}$

When P =2.5 D,

$F=\frac{1}{2.5}=\frac{2}{5}$

When P increases to 2.6 D ,

$F^{\text{'}}=\frac{1}{2.6}=\frac{5}{13}$

Relative decrease in focal length:

$\frac{F-F^{\text{'}}}{F}=\frac{\frac{2}{5}-\frac{5}{13}}{\frac{2}{5}}$

Simplifying:

$=\frac{26-25}{26}=\frac{1}{26}\approx 0.04$

the correct answer is 0.04 .

The apparent depth ($d_a$) is given by the formula:

$d_a=\frac{d}{n}$​

where $d$is the actual depth of the coin and $n$ is the refractive index of the medium (water in this case).

$d_a=\frac{H}{\frac{4}{3}}$​

To solve this, multiply $H$ by the reciprocal of $\frac{4}{3}$:

$d_a=H\times \frac{3}{4}=\frac{3H}{4}$​

To find the radius of curvature ($R$) of the convex refracting surface, we use the refraction formula at a spherical surface:

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$​​

where:

• $n_1 = 1.3$ (refractive index of medium A),
• $n_2 = 1.4$ (refractive index of medium B),
• $u = -26$ (object distance, taken as negative for convex surface),
• $v = \infty$ (image at infinity).

$$\frac{1.4}{\infty} - \frac{1.3}{-26} = \frac{1.4 - 1.3}{R}$$​

Since $\frac{1.4}{\infty} = 0$, the equation simplifies to:

$$\frac{1.3}{26} = \frac{0.1}{R}$$

​ $R=\frac{0.1\times 26}{1.3}=\frac{2.6}{1.3}=2\text{ cm}$

$\boxed{A:2\text{ cm}}$​

To determine the nature of the image formed by a convex spherical surface separating air and glass, we need to consider the refraction of light at the convex surface. The nature of the image (real or virtual) depends on the position of the object relative to the focal point of the convex surface.

Identify the given parameters: object distance (\(u = -4R\)), radius of curvature (\(R\)), and the refractive indices of air (\(n1 = 1\)) and glass (\(n2 > 1\)).

Use the refraction formula for a spherical surface:
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\)
where \(u\) is the object distance, \(v\) is the image distance, and \(R\) is the radius of curvature.

Substitute the given values into the formula:
\(\frac{n_2}{v} - \frac{1}{-4R} = \frac{n_2 - 1}{R}\)

Solve for the image distance \(v\). As the object moves towards the surface (\(u\) increases from \(-4R\) to \(0\)), analyze how the image distance \(v\) changes. Initially, when the object is far (\(u = -4R\)), the image is virtual. As the object moves closer to the surface, the image becomes real when the object is within the focal length of the convex surface.

Final Answer:
D. first real and then virtual

$$\begin{gathered} \text{ (b) }{P}_{ij}=2P_t+P \\ -\frac{1}{f_Q}=\frac{2}{f_{\ell }}-\frac{1}{f_m} \\ =\frac{4(\mu -1)}{R}-\frac{2}{-R}=\frac{1}{R}(4\mu -4+2) \\ -\frac{1}{f_{eq}}=\frac{1}{R}(4\mu -2) \\ \Rightarrow \frac{1}{f_{eq}}=\frac{-1}{R}(4\mu -2) \\ f_{eq}=\frac{R}{2} \\ R=2f_{eq}=2\left(\frac{R}{4\mu -2}\right)=\frac{R}{(2\mu -1)} \end{gathered}$$

To determine the nature of the image formed by a convex spherical surface separating air and glass, we need to consider the refraction of light at the convex surface. The nature of the image (real or virtual) depends on the position of the object relative to the focal point of the convex surface.

Identify the given parameters: object distance (\(u = -4R\)), radius of curvature (\(R\)), and the refractive indices of air (\(n1 = 1\)) and glass (\(n2 > 1\)).

Use the refraction formula for a spherical surface:
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\)
where \(u\) is the object distance, \(v\) is the image distance, and \(R\) is the radius of curvature.

Substitute the given values into the formula:
\(\frac{n_2}{v} - \frac{1}{-4R} = \frac{n_2 - 1}{R}\)

Solve for the image distance \(v\). As the object moves towards the surface (\(u\) increases from \(-4R\) to \(0\)), analyze how the image distance \(v\) changes. Initially, when the object is far (\(u = -4R\)), the image is virtual. As the object moves closer to the surface, the image becomes real when the object is within the focal length of the convex surface.

Final Answer:
D. first real and then virtual

For dispersion without deviation,

$$\begin{gathered} {\mathrm{\delta }}_{\mathrm{net}} = {\mathrm{\delta }}_1 + {\mathrm{\delta }}_2 \\ \left({\mathrm{\mu }}_1 - 1\right){\mathrm{A}}_1 + \left({\mathrm{\mu }}_2 - 1\right){\mathrm{A}}_2 = 0 \\ {\mathrm{A}}_2 = -\frac{\left({\mathrm{\mu }}_1 - 1\right){\mathrm{A}}_1}{\left({\mathrm{\mu }}_2 - 1\right)} \\ {\mathrm{A}}_2 = -\frac{\left(1.54 - 1\right)\times 4}{\left(1.72- 1\right)} = -3° \end{gathered}$$

The negative dign shows that P₂ must be oriented in opposite direction of P₁.

• Using the lens maker's formula: $$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right).$$
• In air ($n_{\text{medium}} = 1$): $$\frac{1}{f_1} = \left( \frac{1.5}{1} - 1 \right) \frac{1}{2} = \frac{1}{4}, \quad f_1 = 4 \, \text{cm}.$$
• In a medium with refractive index 1.2: $$n_{\text{relative}} = \frac{1.5}{1.2} = 1.25, \quad \frac{1}{f_2} = (1.25 - 1) \frac{1}{2} = \frac{1}{8}, \quad f_2 = 8 \, \text{cm}.$$
• Ratio: $$\frac{f_1}{f_2} = \frac{4}{8} = \frac{1}{2}.$$

Thus, $\frac{f_1}{f_2} = \frac{1}{2}$.

We are asked to find the angle of the prism ($A$) when:

1. The angle of the prism is equal to the angle of minimum deviation ($D_m = A$).
2. The refractive index of the prism is $\mu = \sqrt{3}$

The relationship between the refractive index $\mu$, the angle of the prism $A$, and the angle of minimum deviation $D_m$​ is:

$$\mu = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$​

Since $D_m = A$substitute $D_m = A$ into the equation:

$$\mu = \frac{\sin\left(A\right)}{\sin\left(\frac{A}{2}\right)}$$​

Substitute $\mu = \sqrt{3}$​ into the equation:

$$\sqrt{3} = \frac{\sin\left(A\right)}{\sin\left(\frac{A}{2}\right)}$$​

1. Simplify the equation:

   $$\sin(A) = \sqrt{3} \cdot \sin\left(\frac{A}{2}\right)$$

2. Use the double-angle identity:

   $$\sin(A) = 2 \cdot \sin\left(\frac{A}{2}\right) \cdot \cos\left(\frac{A}{2}\right)$$

   Substitute $\sin(A)$sin(A) into the equation:

   $$2 \cdot \sin\left(\frac{A}{2}\right) \cdot \cos\left(\frac{A}{2}\right) = \sqrt{3} \cdot \sin\left(\frac{A}{2}\right)$$

3. Divide through by $\sin\left(\frac{A}{2}\right)$ (assuming $\sin\left(\frac{A}{2}\right) \neq 0$):

   $$2 \cdot \cos\left(\frac{A}{2}\right) = \sqrt{3}$$

4. Solve for $\cos\left(\frac{A}{2}\right)$:

   $$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$$​

5. From the standard trigonometric values:

   $$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2} \implies \frac{A}{2} = 30^\circ$$

6. Solve for $A$:

   $$A = 60^\circ$$

The angle of the prism is:

$$\boxed{60^\circ}$$​

Correct Option: A. $60^\circ$

$\sin {\theta }_{1C}=\frac{n_1}{n_2}$

$\sin {\theta }_{2C}=\frac{n_1}{n_3}$

$\frac{\sin {\theta }_{1C}}{\sin {\theta }_{2\text{C}}}=\frac{n_3}{n_2}$

$\sin {\theta }_{1C}-\sin {\theta }_{2C}=\frac{1}{2}$

$\Rightarrow \sin {\theta }_{1C}-\frac{2}{5}\sin {\theta }_{1C}=\frac{1}{2}$

$\sin {\theta }_{1C}=\frac{5}{6}$

${\theta }_{1C}={\sin }^{-1}\frac{5}{6}$