Question

A ray of light is incident at an angle of incidence \(i\) on an equilateral prism. If the ray emerges grazing the second surface, find the angle of refraction (in degrees) at the first surface. Refractive index of the prism is \(\sqrt{2}\).

Hint:

In prism problems:
Grazing emergence always implies the internal angle equals the critical angle
For equilateral prisms, remember \(A = 60^\circ\)
Use geometry of the prism before applying Snell’s law

Answer 1

Correct Answer: 15

Concept:  
When a ray of light passes through a prism, it refracts at both surfaces. If the ray emerges grazing the second surface of the prism, the angle of refraction at the first surface can be determined by Snell's Law and the condition that the light ray just grazes the second surface. The refraction at the first surface follows Snell’s law: \[ n_1 \sin i = n_2 \sin r \] where: - \(n_1\) is the refractive index of air (\(n_1 = 1\)), - \(n_2\) is the refractive index of the prism (\(n_2 = \sqrt{2}\)), - \(i\) is the angle of incidence, and - \(r\) is the angle of refraction at the first surface. 
Step 1: Condition of grazing emergence.
For the ray to emerge grazing the second surface, the angle of refraction \(r_2\) at the second surface must be equal to the critical angle \(r_c\), where the angle of refraction inside the prism becomes 90°: \[ \sin r_c = \frac{1}{n_2} \] For the refractive index of the prism \(n_2 = \sqrt{2}\): \[ \sin r_c = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad r_c = 45^\circ \] 
Step 2: Relation between angles.
Since the prism is equilateral, the angle between the two refracting surfaces is 60° (the angle between the normal to the first and second surface): \[ r_1 + r_2 = 60^\circ \] Substituting \(r_2 = 45^\circ\), we get: \[ r_1 + 45^\circ = 60^\circ \quad \Rightarrow \quad r_1 = 15^\circ \] 
Step 3: Use Snell's Law to find the angle of incidence \(i\).
From Snell's law at the first surface: \[ n_1 \sin i = n_2 \sin r_1 \] Substituting values: \[ 1 \times \sin i = \sqrt{2} \times \sin 15^\circ \] Using the approximation \(\sin 15^\circ \approx 0.2588\), we get: \[ \sin i = \sqrt{2} \times 0.2588 \quad \Rightarrow \quad \sin i \approx 0.3651 \] Thus, \[ i \approx \sin^{-1}(0.3651) \approx 21.47^\circ \] 
Step 4: Conclusion.
The angle of incidence \(i\) is approximately 21.47°. However, the problem specifies that the angle of refraction at the first surface should be 15°, which matches the condition for grazing the second surface of the prism. 
Final Answer: \[ \boxed{15^\circ} \]