Question

A ray of light is incident on the face of a triangular prism of refracting angle $60^\circ$ and it just suffers total internal reflection at the other face. Find the angle of incidence for the ray if the refractive index of the material of the prism is $\sqrt{2}$.

Hint:

The specific phrasing "just suffers TIR" or "grazes the emergent surface" invariably implies exactly one thing mathematically: $r_2 = C$.

Answer 1

Concept:
• When a ray of light cleanly traverses a triangular optical prism, it undergoes refraction strictly at two distinct interfaces: upon entering the first face and upon exiting the second face.

• The fundamental geometric relation deeply connecting the two internal angles of refraction ($r_1$ and $r_2$) to the prism's characteristic refracting angle $A$ is universally given by $A = r_1 + r_2$.

• The phrase "just suffers total internal reflection" acts as a critical physical trigger word, implying that the light ray hits the second internal boundary exactly at the critical angle, meaning the final angle of emergence is a grazing $90^\circ$.

• The critical angle $C$ for any transparent medium located in air is rigorously defined by the sine relation: $\sin C = \frac{1}{\mu}$.

Step 1: Identify the given optical parameters
The principal refracting angle of the triangular prism is explicitly $A = 60^\circ$.
The absolute refractive index of the glass material is strictly $\mu = \sqrt{2}$.
The emergent condition "just suffers total internal reflection" mathematically mandates that the internal angle of incidence squarely on the second face is exactly equal to the critical angle ($r_2 = C$).

Step 2: Calculate the critical angle for the prism material
We systematically use the critical angle formula to find the exact value of $C$:
\[ \sin C = \frac{1}{\mu} \]
Substitute the known refractive index securely into the equation:
\[ \sin C = \frac{1}{\sqrt{2}} \]
From foundational trigonometry, we universally recognize this specific sine value:
\[ C = 45^\circ \]
Therefore, the light ray strikes the second internal face at precisely $r_2 = 45^\circ$.

Step 3: Calculate the angle of refraction at the first face
We leverage the core geometric prism identity to meticulously find the first internal refraction angle $r_1$:
\[ A = r_1 + r_2 \]
Substitute the known values for the prism angle and the derived second refraction angle:
\[ 60^\circ = r_1 + 45^\circ \]
Cleanly isolate $r_1$:
\[ r_1 = 60^\circ - 45^\circ = 15^\circ \]
This indicates that the incident light ray bends internally to a $15^\circ$ angle immediately after penetrating the first face.

Step 4: Calculate the initial angle of incidence using Snell's Law
We rigorously apply Snell's Law specifically at the very first air-glass interface to discover the initial angle of incidence $i_1$:
\[ \mu_1 \sin i_1 = \mu_2 \sin r_1 \]
Since the ray originates externally from ambient air, we establish $\mu_1 = 1$, and the prism's index is $\mu_2 = \sqrt{2}$:
\[ 1 \times \sin i_1 = \sqrt{2} \times \sin(15^\circ) \]
To strictly evaluate $\sin(15^\circ)$ without a calculator, we expand it strategically using standard angle subtraction formulas:
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]
Substitute the standard known trigonometric exact values:
\[ \sin(15^\circ) = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) \]
\[ \sin(15^\circ) = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]
Now, carefully substitute this exact expanded expression squarely back into our pending Snell's Law equation:
\[ \sin i_1 = \sqrt{2} \times \left( \frac{\sqrt{3} - 1}{2\sqrt{2}} \right) \]
The $\sqrt{2}$ terms wonderfully and cleanly cancel out completely:
\[ \sin i_1 = \frac{\sqrt{3} - 1}{2} \]

Step 5: Conclusion
We can cleanly express the final exact incident angle formally using the inverse sine function:
\[ i_1 = \sin^{-1}\left(\frac{\sqrt{3} - 1}{2}\right) \]
(For context, since $\sqrt{3} \approx 1.732$, $\sin i_1 \approx 0.366$, which rigorously corresponds to an angle of roughly $21.47^\circ$).