A light ray enters through a right angled prism at point P with the angle of incidence 30° as shown in figure. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is:

Based on the provided figure, the prism is a right-angled prism with the right angle at vertex $A$ (so the angle of the prism is $A=90°$). The light ray enters face $AB$ at point $P$ with an angle of incidence $i=30°$ and emerges along the face $AC$, which means the angle of emergence is $e=90°$.

Let the refractive index of the prism be $\mu$, the angle of refraction at the first face $AB$ be $r_1$, and the angle of incidence at the second face $AC$ be $r_2$.

Step 1: Apply Snell's Law at the first face (AB)
The ray enters from air (refractive index = 1) into the prism of refractive index $\mu$:
$1\cdot \sin \left(i\right)=\mu \cdot \sin \left(r_1\right)$
Substituting $i=30°$:
$\sin (30°)=\mu \sin \left(r_1\right)$
$\frac{1}{2}=\mu \sin \left(r_1\right)$
$\sin \left(r_1\right)=\frac{1}{2\mu }$

Step 2: Relate the angles inside the prism
For a prism with refracting angle $A$, the relationship between internal angles is:
$r_1+r_2=A$
Since the prism is right-angled at vertex $A$, we have $A=90°$:
$r_1+r_2=90°$
Therefore, the angle of incidence at the second face is:
$r_2=90°-r_1$

Step 3: Apply Snell's Law at the second face (AC)
The ray emerges along the face $AC$, which means the angle of refraction in air is $e=90°$:
$\mu \cdot \sin \left(r_2\right)=1\cdot \sin (90°)$
$\mu \sin (90°-r_1)=1$
Using the trigonometric identity $\sin (90°-\theta )=\cos \left(\theta \right)$:
$\mu \cos \left(r_1\right)=1$
$\cos \left(r_1\right)=\frac{1}{\mu }$

Step 4: Solve for the refractive index ($\mu$)
Using the fundamental trigonometric identity:
${\sin }^2\left(r_1\right)+{\cos }^2\left(r_1\right)=1$
Substitute the values from Steps 1 and 3:
${\left(\frac{1}{2\mu }\right)}^2+{\left(\frac{1}{\mu }\right)}^2=1$
$\frac{1}{4{\mu }^2}+\frac{1}{{\mu }^2}=1$
Taking $4{\mu }^2$ as the common denominator:
$\frac{1+4}{4{\mu }^2}=1$
$\frac{5}{4{\mu }^2}=1$
$4{\mu }^2=5$
${\mu }^2=\frac{5}{4}$
Taking the square root on both sides:
$\mu =\frac{\sqrt{5}}{2}$