Question

Consider an equilateral prism (refractive index \( \sqrt{2} \)). A ray of light is incident on its one surface at a certain angle \( i \). If the emergent ray is found to graze along the other surface, then the angle of refraction at the incident surface is close to

• A. \(15^\circ\)
• B. \(40^\circ\)
• C. \(20^\circ\)
• D. \(30^\circ\)

Hint:

When a ray grazes the surface during emergence, the angle of refraction at that surface equals the critical angle.

Answer 1

The Correct Option is A

To solve this problem, we need to analyze the refraction of light through an equilateral prism with a refractive index \( \sqrt{2} \) and determine the angle of refraction at the incident surface when the emergent ray grazes along the second surface. Let's break down the solution into steps:

Step-by-Step Solution:

Understanding the Setup: We have an equilateral prism, meaning all angles are \(60^\circ\). The refractive index of the prism is \( \sqrt{2} \). A light ray is incident on one of its faces at an angle \( i \).

Condition for Grazing Emergence: For a ray to graze along the surface, the angle of emergence must be \(90^\circ\). By applying Snell's Law at the emergent surface, where the angle of refraction is \(90^\circ\), we get:

\(n \sin(r_2) = 1 \cdot \sin(90^\circ)\)

Since \(\sin(90^\circ) = 1\), then \(n \sin(r_2) = 1\).

Thus, \(\sin(r_2) = \frac{1}{n} = \frac{1}{\sqrt{2}}\).

This implies \( r_2 = 45^\circ \).

Application of Geometry of the Prism: For an equilateral prism:

\(A + \delta = (r_1 + r_2)\) and \( A = 60^\circ \) because it's an equilateral prism.

Thus, \(60^\circ = r_1 + 45^\circ \), which solves to \( r_1 = 15^\circ \).

Conclusion: The angle of refraction \( r_1 \) at the incident surface is \(15^\circ\).

Verification with Snell's Law:

Finally, we can verify \( r_1 = 15^\circ \) using Snell's Law:

\(n_1 \sin(i) = n_2 \sin(r_1)\), where \(n_1 = 1\) (air) and \(n_2 = \sqrt{2}\).

So, \(\sin(i) = \sqrt{2} \cdot \sin(15^\circ)\). But since exact \( i \) was not asked, the question verifies for \( r_1 \).

Therefore, the correct answer is \(15^\circ\).