Two equilateral-triangular prisms P₁ and P₂ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism P₁ at an angle of incidence 𝜃 such that the outgoing ray undergoes minimum deviation in prism P₂. If the respective refractive indices of P₁ and P₂ are $\sqrt{\frac{3}{2}}$ and √3, then $\mathrm{\theta <!--\; \theta \; -->}={\sin }^{-<!--\; -\; -->1}<!--\; \; -->[\sqrt{\frac{3}{2}}\sin <!--\; \; -->\left(\frac{\mathrm{\pi <!--\; \pi \; -->}}{\mathrm{\beta <!--\; \beta \; -->}}\right)]$, where the value of β is ______.

The correct answer is 12.

Step 1: Understanding the Geometry from the Figure
As shown in the figure, two equilateral-triangular prisms, P₁ and P₂, are placed adjacent to each other in vacuum. The right refracting face of P₁ is parallel to the left refracting face of P₂. A ray of light enters P₁ at an angle of incidence 𝜃, refracts through P₁, exits into the vacuum region between the two prisms, and then enters P₂.

Step 2: Condition for Minimum Deviation in P₂
Since P₂ is an equilateral-triangular prism, its prism angle is:
$A_2={60}^{\circ }$
The refractive index of P₂ is given as:
${\mu }_2=\sqrt{3}$
For a light ray undergoing minimum deviation in an equilateral prism, the angle of refraction inside the prism at both refracting surfaces is equal and given by:
$r=\frac{A_2}{2}={30}^{\circ }$
Let $i_2$ be the angle of incidence at the first face (left face) of P₂. Applying Snell's law at this surface:
$1·\sin \left(i_2\right)={\mu }_2\sin \left({30}^{\circ }\right)$
Substituting the value of ${\mu }_2$:
$\sin \left(i_2\right)=\sqrt{3}·\frac{1}{2}=\frac{\sqrt{3}}{2}$
Therefore, the angle of incidence at P₂ is:
$i_2={60}^{\circ }$

Step 3: Finding the Angle of Emergence from P₁
Because the right face of P₁ is parallel to the left face of P₂, the normal to the right face of P₁ is parallel to the normal to the left face of P₂. Thus, the angle of emergence $e_1$ of the ray leaving P₁ must be equal to the angle of incidence $i_2$ at P₂:
$e_1=i_2={60}^{\circ }$

Step 4: Analyzing the Refraction in P₁
The refractive index of P₁ is given as:
${\mu }_1=\sqrt{\frac{3}{2}}$
Let $r_2$ be the angle of refraction inside P₁ at its second face (the right face). Applying Snell's law at this interface:
${\mu }_1\sin \left(r_2\right)=1·\sin \left(e_1\right)$
Substituting the known values:
$\sqrt{\frac{3}{2}}\sin \left(r_2\right)=\sin \left({60}^{\circ }\right)=\frac{\sqrt{3}}{2}$
Solving for $\sin \left(r_2\right)$:
$\sin \left(r_2\right)=\frac{\sqrt{3}}{2}·\sqrt{\frac{2}{3}}=\frac{1}{\sqrt{2}}$
This gives:
$r_2={45}^{\circ }$

Step 5: Finding the Angle of Refraction at the First Face
Since P₁ is also an equilateral-triangular prism, its prism angle is $A_1={60}^{\circ }$. Using the relationship between the internal refraction angles:
$r_1+r_2=A_1$
Substituting $r_2={45}^{\circ }$ and $A_1={60}^{\circ }$:
$r_1={60}^{\circ }-{45}^{\circ }={15}^{\circ }$

Step 6: Calculating the Angle of Incidence 𝜃
Applying Snell's law at the first refracting surface of P₁:
$1·\sin \left(\theta \right)={\mu }_1\sin \left(r_1\right)$
$\sin \left(\theta \right)=\sqrt{\frac{3}{2}}\sin \left({15}^{\circ }\right)$
Expressing ${15}^{\circ }$ in radians:
${15}^{\circ }=\frac{\pi }{12}$
Substituting this back, we obtain:
$\theta ={\sin }^{-1}\left[\sqrt{\frac{3}{2}}\sin \left(\frac{\pi }{12}\right)\right]$
Comparing this with the given template:
$\theta ={\sin }^{-1}\left[\sqrt{\frac{3}{2}}\sin \left(\frac{\pi }{\beta }\right)\right]$
We find:
$\beta =12$