Question

When a glass prism of refracting angle \(60^\circ\) is immersed in a liquid its angle of minimum deviation is \(30^\circ\). The critical angle of glass with respect to the liquid medium is:

• A. \(45^\circ\)
• B. \(42^\circ\)
• C. \(50^\circ\)
• D. \(52^\circ\)

Hint:

Critical angle: \( \sin C = \frac{1}{\mu} \). Always find refractive index first.

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the angle of minimum deviation, the prism, and the critical angle of glass with respect to the liquid. This involves Snell's Law and the concept of refraction through the prism.

1. The given prism has a refracting angle \( A = 60^\circ \) and the angle of minimum deviation \( \delta_m = 30^\circ \).
2. The minimum deviation occurs when the refracted ray inside the prism is symmetrical with respect to the face of the prism. For a prism, the formula relating the angle of minimum deviation, prism angle, and refractive index is: \(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
3. Substituting the given values \( A = 60^\circ \) and \( \delta_m = 30^\circ \) into the formula: \(n = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}\)
4. The trigonometric values are: \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\) and \(\sin(30^\circ) = \frac{1}{2}\).
5. Plugging in these values: \(n = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}\).
6. Now, the critical angle \( C \) for the glass-liquid interface can be found using the relation: \(\sin(C) = \frac{1}{n}\), since refractive index from glass to liquid is given as the reciprocal.
7. So, \(\sin(C) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\), which means: \(C = 45^\circ\).

The correct answer is, therefore, \( 45^\circ \), which matches option 1.