Q.At what angle should a ray of light be incident on the face of a prism of refracting angle $60^\circ$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$.

Concept: Total Internal Reflection (TIR) — the ray must strike the second face at an angle equal to or greater than the critical angle.

Step 1: Find the critical angle for the prism-air interface.

For TIR at the second face, the angle of incidence inside the prism at that face must equal the critical angle $C$:

Step 2: Relate prism angles.

For a prism with refracting angle $A = 60^\circ$, the geometry inside gives:

where $r_1$ is the angle of refraction at the first face and $r_2$ is the angle of incidence at the second face. For TIR at the second face, $r_2 = C = 41^\circ$. Hence: …

The key idea here is that "just suffers total internal reflection" means the ray hits the second face at exactly the critical angle — any less and it would refract out, any more and it would still reflect but we want the threshold. This is a classic prism problem that ties together Snell's law at the first face and the critical angle condition at the second face, linked by the prism's geometry.

Let’s unpack why this works. Total internal reflection (TIR) happens when light tries to go from a denser medium (prism, $n = 1.524$) to a rarer medium (air, $n = 1$) at an angle greater than the critical angle. The critical angle $C$ is given by $\sin C = \frac{1}{n}$. For the ray to "just" suffer TIR, the angle of incidence on the second face must equal $C$. The prism angle $A = 60^\circ$ then relates this internal angle to the angle of refraction at the first face through simple geometry.

Now let’s work through it step by step.

1. Find the critical angle for the prism material. The refractive index of the prism relative to air is $n = 1.524$. For TIR at the second face, the angle of incidence inside the prism on that face must be at least the critical angle $C$, where

So

Calculating: $\frac{1}{1.524} \approx 0.6562$, so $C \approx 41.0^\circ$ (more precisely, $41.0^\circ$ is a good approximation; let's keep it exact for now).

2. Relate the prism geometry to the internal angles. In a prism, the sum of the angles inside the triangle formed by the ray and the two faces is $180^\circ$. Specifically, if $r_1$ is the angle of refraction at the first face (inside the prism) and $r_2$ is the angle of incidence on the second face (also inside the prism), then for a prism of angle $A$:

This is because the normals at the two faces are inclined at angle $A$, and the ray's path inside forms a triangle with them.

Here $A = 60^\circ$, and for "just" TIR, $r_2 = C \approx 41.0^\circ$. Therefore:

So $r_1 \approx 60^\circ - 41.0^\circ = 19.0^\circ$.

3. Apply Snell's law at the first face to find the external angle of incidence $i$. At the first face, light goes from air ($n = 1$) into the prism ($n = 1.524$). Snell's law:

Substituting $r_1 = 60^\circ - C$:

Method: Critical Angle & Prism Geometry Approach

This problem uses the condition for total internal reflection (TIR) combined with Snell’s law and the angle-sum relation in a prism.

Step 1 — Find the critical angle for the prism material

For TIR at the second face, the angle of incidence inside the prism at that face must equal the critical angle $C$.

From Snell’s law:

Given $\mu = 1.524$:

Key result: The ray inside the prism strikes the second face at $41.0^\circ$ for TIR.

Step 2 — Relate prism angles

In a prism with refracting angle $A = 60^\circ$, the angles inside the prism are related by:

Here $r_2$ is the angle of incidence at the second face, which equals $C$ for TIR:

Thus:

Step 3 — Apply Snell’s law at the first face …

Here are the most common mistakes students make when solving this Total Internal Reflection (TIR) problem, along with how to avoid each.

Mistake 1: Confusing the angle of incidence at the first face with the angle at the second face

The error:

Students often think the given condition (“just suffers TIR at the other face”) refers to the angle of incidence at the first face. They then try to directly apply Snell’s law at the first surface using $1.524$ and $60^\circ$, leading to a wrong answer.

How to avoid:

• Read carefully: The TIR happens at the second face (the other face).
• The angle of incidence at the second face ($r_2$) must equal the critical angle for the prism material.
• First find the critical angle $C$ using:

This $C$ is the value of $r_2$ at the second face.

Mistake 2: Forgetting the prism geometry relation between $r_1$ and $r_2$

The error:

After finding $r_2 = C$, students sometimes forget that for a prism, the sum of the two refraction angles equals the prism angle $A$:

They might incorrectly set $r_1 = A$ or $r_2 = A$.

How to avoid:

• Always write the prism relation first:

• Here $A = 60^\circ$ and $r_2 = C \approx 41^\circ$, so:

Mistake 3: Using the wrong Snell’s law sign or direction at the first face

The error:

When applying Snell’s law at the first face, students sometimes write:

but forget that the ray is going from air ($\mu = 1$) into prism ($\mu = 1.524$). They may invert the fraction.

How to avoid:

• Always write Snell’s law in the correct order:

• Since $\mu_{\text{air}} = 1$, this becomes:

• Calculate $\sin 19^\circ \approx 0.3256$, then:

Mistake 4: Not checking if the calculated $i$ is physically possible

The error:

Students sometimes get $\sin i > 1$ due to a calculation slip (e.g., using $r_1 = 41^\circ$ instead of $19^\circ$). They then report an impossible angle.

How to avoid:

• Always check: $\sin i$ must be $\leq 1$. …