$\text{Two equilateral-triangular prisms}{P}_1\text{and}{P}_2\text{are kept with their sides parallel to each other, in vacuum,}$

$\text{as shown in the figure. }\text{A light ray enters prism}{\mathrm{ P}}_1\text{at an angle of incidence}\theta \text{such that the outgoing ray}$

$\text{undergoes minimum deviation in prism}{\mathrm{ P}}_2.\text{If the respective refractive indices of}{\mathrm{ P}}_1\text{and}{\mathrm{ P}}_{\mathrm{2 }}$

${\mathrm{}}_{}\text{are}\sqrt{\frac{3}{2}}\text{and}\sqrt{3},\text{then}\mathrm{ \theta }={\sin }^{-1}\left[\sqrt{\frac{3}{2}}\sin \right(\frac{\pi }{\beta }\left)\right],\text{where the value of}\mathrm{ \beta }\text{ is \_\_\_\_\_}$

The correct answer is 12.

1. Understanding the Given Parameters:
We are given two equilateral triangular prisms, _{$P_1$} and _{$P_2$}. Since both are equilateral prisms, their refracting angles are:
$A={60}^{°}=\frac{\pi }{3}\text{rad}$

The refractive indices of the prisms are:
Refractive index of _{$P_1$}: ${\mu }_1=\sqrt{\frac{3}{2}}$
Refractive index of _{$P_2$}: ${\mu }_2=\sqrt{3}$

2. Minimum Deviation in Prism _{$P_2$}:
For a ray undergoing minimum deviation inside an equilateral prism of refractive index _{${\mu }_2$}, the angle of refraction inside the prism is:
$r_2^{\text{'}}=\frac{A}{2}=\frac{{60}^{°}}{2}={30}^{°}$

Using Snell's law at the first surface of prism _{$P_2$} (where the incident ray comes from vacuum into _{$P_2$}):
$1\cdot \sin \left(i_2\right)={\mu }_2\sin \left(r_2^{\text{'}}\right)$

Substitute _{${\mu }_2=\sqrt{3}$} and _{$r_2^{\text{'}}={30}^{°}$}:
$\sin \left(i_2\right)=\sqrt{3}\cdot \sin \left({30}^{°}\right)=\sqrt{3}\cdot \frac{1}{2}=\frac{\sqrt{3}}{2}$

Thus, the angle of entry into prism _{$P_2$} is:
$i_2={60}^{°}$

3. Ray Trajectory between Prism _{$P_1$} and _{$P_2$}:
The right face of _{$P_1$} and the left face of _{$P_2$} are parallel to each other. Therefore, the angle of emergence _{$e_1$} from the second face of prism _{$P_1$} must equal the angle of incidence _{$i_2$} on prism _{$P_2$}:
$e_1=i_2={60}^{°}$

4. Refraction inside Prism _{$P_1$}:
Using Snell's law at the emergent face of _{$P_1$}:
${\mu }_1\sin \left(r_2\right)=1\cdot \sin \left(e_1\right)$

Substitute _{${\mu }_1=\sqrt{\frac{3}{2}}$} and _{$e_1={60}^{°}$}:
$\sqrt{\frac{3}{2}}\sin \left(r_2\right)=\sin \left({60}^{°}\right)=\frac{\sqrt{3}}{2}$

Solving for _{$\sin \left(r_2\right)$}:
$\sin \left(r_2\right)=\frac{\sqrt{3}}{2}\cdot \sqrt{\frac{2}{3}}=1\sqrt{2}$

Hence, the angle of refraction _{$r_2$} at the second surface of _{$P_1$} is:
$r_2={45}^{°}$

For prism _{$P_1$}, the relation between the refracting angle _$A$ and internal angles of refraction _{$r_1$} and _{$r_2$} is:
$r_1+r_2=A$

Substitute _{$A={60}^{°}$} and _{$r_2={45}^{°}$}:
$r_1={60}^{°}-{45}^{°}={15}^{°}=\frac{\pi }{12}\text{rad}$

5. Finding the Angle of Incidence _$\theta$:
Using Snell's law at the first surface of _{$P_1$}:
$1\cdot \sin \left(\theta \right)={\mu }_1\sin \left(r_1\right)$

Substitute _{${\mu }_1=\sqrt{\frac{3}{2}}$} and _{$r_1=\frac{\pi }{12}$}:
$\sin \left(\theta \right)=\sqrt{\frac{3}{2}}\sin \left(\frac{\pi }{12}\right)$

Taking the inverse sine of both sides:
$\theta ={\sin }^{-1}\left[\sqrt{\frac{3}{2}}\sin \right(\frac{\pi }{12}\left)\right]$

Comparing this expression with the given form:
$\theta ={\sin }^{-1}\left[\sqrt{\frac{3}{2}}\sin \right(\frac{\pi }{\beta }\left)\right]$

We find that:
$\beta =12$