Find the value of the angle of emergence from the prism. Refractive index of the glass is square root of 3.

The correct answer is 60⁰.

Step-by-Step Derivation:

1. Identify the angles of the prism:
From the given image, the prism is a right-angled triangle.
The angle at the bottom-left vertex is $90°$, and the angle at the bottom-right vertex is $60°$.
Therefore, the angle at the top vertex (let's call it $A$) is:
$A=180°-(90°+60°)=30°$

2. Determine the path of the light ray inside the prism:
The incident light ray enters normally (perpendicularly) to the hypotenuse face of the prism.
Since the angle of incidence at this first surface is $0°$, the ray passes straight into the prism without any deviation.

3. Calculate the angle of incidence at the vertical face:
Let $P$ be the point of entry on the hypotenuse, and $Q$ be the point where the ray hits the vertical face.
In the right-angled triangle formed by the top vertex $A$, the entry point $P$, and the point $Q$ on the vertical face ($\mathrm{\Delta APQ}$):
The angle $\angle \mathrm{APQ}=90°$ (since the ray enters normally).
The angle at the top vertex is $\angle \mathrm{PAQ}=30°$.
Therefore, the third angle in the triangle, $\angle \mathrm{AQP}$, is:
$\angle \mathrm{AQP}=180°-(90°+30°)=60°$
The normal to the vertical face is perpendicular ($90°$) to that face. Thus, the angle of incidence ($i$) on the vertical face is:
$i=90°-60°=30°$

4. Apply Snell's Law to find the angle of emergence ($e$):
According to Snell's Law at the interface of the vertical face:
$\mu ·sin\left(i\right)={\mu }_{\mathrm{air}}·sin\left(e\right)$
Where:
- The refractive index of glass is $\mu =\sqrt{3}$
- The refractive index of air is ${\mu }_{\mathrm{air}}=1$
- The angle of incidence is $i=30°$
Substituting these values:
$\sqrt{3}·sin(30°)=1·sin\left(e\right)$
Since $sin(30°)=\frac{1}{2}$:
$sin\left(e\right)=\frac{\sqrt{3}}{2}$
Taking the inverse sine:
$e=60°$

Thus, the angle of emergence from the prism is 60⁰.