Question

Prove that total internal reflection of light ray from refracting surface is possible only when the value of angle of prism A be more than \(\sin^{-1}\left(\frac{1}{n}\right)\), where 'n' is refractive index of the material of prism.

Hint:

This principle is fundamental to the design of optical instruments. For example, 45\(^\circ\)-90\(^\circ\)-45\(^\circ\) prisms are used in binoculars and periscopes. For typical glass (\(n \approx 1.5\)), the critical angle is \(C \approx 42^\circ\). Since \(45^\circ > 42^\circ\), these prisms are designed to produce total internal reflection.

Answer 1

Step 1: Understanding the Concept:
Total Internal Reflection (TIR) is a phenomenon where a light ray traveling from a denser medium (e.g., prism with refractive index \(n\)) to a rarer medium (e.g., air with refractive index \(\approx 1\)) is completely reflected back into the denser medium. This occurs only if the angle of incidence in the denser medium is greater than a specific angle called the critical angle, C. The question asks for the condition on the prism's apex angle A for TIR to be possible at the second face.

Step 2: Key Formula or Approach:
1. Critical Angle (C): The critical angle is defined by Snell's Law when the angle of refraction is 90\(^\circ\). For a prism in air, \( n \sin(C) = 1 \sin(90^\circ) \), which gives \(\sin(C) = \frac{1}{n}\) or \(C = \sin^{-1}\left(\frac{1}{n}\right)\).
2. Condition for TIR: For TIR to occur at the second face of the prism, the angle of incidence there, \(r_2\), must be greater than or equal to the critical angle C. That is, \(r_2 \ge C\).
3. Prism Angle Formula: In a prism, the apex angle \(A\) is related to the angle of refraction at the first face (\(r_1\)) and the angle of incidence at the second face (\(r_2\)) by the geometric relation \(A = r_1 + r_2\).

Step 3: Detailed Explanation:
Let a ray of light enter the first face of the prism and refract at an angle \(r_1\). This ray then travels to the second face, striking it at an angle of incidence \(r_2\).
The condition for TIR to take place at this second face is: \[ r_2 \ge C \] Using the prism angle formula, we can express \(r_2\) as \(r_2 = A - r_1\). Substituting this into the TIR condition: \[ A - r_1 \ge C \] Rearranging the inequality to find the condition on A: \[ A \ge C + r_1 \] This inequality shows that for TIR to be possible, the prism angle A must be at least \(C + r_1\). The value of \(r_1\) depends on the initial angle of incidence \(i\) on the first face. To determine if TIR is possible at all, we must check if this condition can be met for any valid \(r_1\).
The minimum possible value for \(r_1\) is 0, which occurs when the light ray enters the prism at normal incidence (\(i = 0\)). If we substitute this minimum value of \(r_1\) into our condition, we get the minimum requirement for A: \[ A \ge C + 0 \] \[ A \ge C \] This means that for TIR to be even a possibility, the prism angle A must be at least equal to the critical angle C. If \(A < C\), then \(A - r_1\) will always be less than C (since \(r_1 \ge 0\)), and TIR can never occur.

Step 4: Final Answer:
Thus, total internal reflection is possible only if the angle of the prism \(A\) is greater than or equal to the critical angle \(C\). Substituting the expression for the critical angle, \(C = \sin^{-1}\left(\frac{1}{n}\right)\), we prove that TIR is possible only when \(A \ge \sin^{-1}\left(\frac{1}{n}\right)\). The question's use of "more than" implies the strict condition for practical observation.