Q.A triangular prism of refracting angle $60^\circ$ is made of a transparent material of refractive index $\dfrac{2}{\sqrt{3}}$. A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.

Concept: Critical Angle Comparison — The ray hits the second face at an angle equal to the critical angle, so it emerges along the face.

1. The ray is incident normally on face KL, so it enters the prism undeviated and strikes face KM at an angle of incidence equal to the prism angle, $i = 60^\circ$.

2. The critical angle for the material is given by

   $\sin \theta_c = \frac{1}{\mu} = \frac{\sqrt{3}}{2} \quad\Rightarrow\quad \theta_c = 60^\circ$.

3. Since $i = \theta_c$, the ray hits the second face at the critical angle. Applying Snell’s law at face KM:

   $\mu \sin 60^\circ = \sin r \;\Rightarrow\; \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{2} = \sin r = 1 \;\Rightarrow\; r = 90^\circ$. …

Why Critical Angle Comparison is the Key

When a ray enters a prism normally, it passes straight through the first face without bending. The real action happens at the second face, where the ray meets the glass-air boundary. The question becomes: does it emerge, or does it reflect internally?

The answer lies in comparing the angle of incidence at the second face with the critical angle for the material. If the incidence angle exceeds the critical angle, total internal reflection occurs. If it's less, the ray emerges with refraction. If it equals the critical angle, the emergent ray grazes along the surface — a special borderline case.

Here, the refractive index $\mu = \frac{2}{\sqrt{3}}$ is unusual enough that the critical angle works out to a clean $60^\circ$ — exactly the prism angle. That's the entire story in a nutshell.

Step-by-Step Solution

1. The ray enters normally on face KL

Normal incidence means the ray strikes perpendicular to the surface. By Snell's law, $\mu_1 \sin i = \mu_2 \sin r$. With $i = 0^\circ$, $\sin 0^\circ = 0$, so $r = 0^\circ$ regardless of the refractive indices. The ray passes straight through the first face without any deviation.

2. Find the angle of incidence at the second face

Inside the prism, the ray travels toward face KM. The prism has a refracting angle $A = 60^\circ$ at vertex K. Since the ray entered normally on face KL, it makes an angle of $0^\circ$ with the normal to KL. Inside the prism, the ray is parallel to the base LM (or nearly so, depending on geometry).

The geometry of a triangular prism tells us: when a ray enters normally on one face, the angle it makes with the normal to the second face equals the prism angle $A$. This is because the normals to the two faces are inclined at the prism angle.

Therefore, at the second face KM, the angle of incidence is:

3. Calculate the critical angle for the material

The critical angle $\theta_c$ is given by:

Substituting $\mu = \frac{2}{\sqrt{3}}$:

Therefore:

4. Compare $i_2$ with $\theta_c$

We have $i_2 = 60^\circ$ and $\theta_c = 60^\circ$. They are exactly equal.

This is the critical case: the ray strikes the second face at precisely the critical angle. It does not undergo total internal reflection (which requires $i > \theta_c$), nor does it emerge normally. Instead, it emerges at the maximum possible angle — grazing along the surface.

5. Apply Snell's law at the second face to find the angle of emergence …