Step 1: Understand the concept of critical angle.
The critical angle is the angle of incidence (in the denser medium) at which the refracted ray travels along the interface, meaning the angle of refraction is exactly $90^\circ$. For angles of incidence greater than the critical angle, total internal reflection occurs and no light passes into the second medium.
Step 2: Identify the media and their refractive indices.
Light travels from medium 1 with refractive index $\mu_1 = 2$ into medium 2 with refractive index $\mu_2 = \sqrt{3}$. Since $\mu_1 > \mu_2$, the first medium is optically denser, so a critical angle exists.
Step 3: Apply Snell's law at the critical angle.
At the critical angle $C$, the angle of refraction is $90^\circ$. Snell's law gives: \[ \mu_1 \sin C = \mu_2 \sin 90^\circ = \mu_2 \] So: \[ \sin C = \frac{\mu_2}{\mu_1} = \frac{\sqrt{3}}{2} \]
Step 4: Find the critical angle.
We know from trigonometry that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Therefore: \[ C = 60^\circ \]
Step 5: Verify the result makes physical sense.
A critical angle of $60^\circ$ means that rays hitting the interface at angles greater than $60^\circ$ (from the normal) will undergo total internal reflection. This is a reasonable critical angle for these two refractive indices.
Step 6: State the final answer.
The critical angle is: \[ \boxed{60^\circ} \]