The exit surface of a prism with refractive index n is coated with a material having refractive index n/2. When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is ____

Prism with coated surface meeting critical angle at minimum deviation

Q. The exit surface of a prism with refractive index \(n\) is coated with a material having refractive index \( \dfrac{n}{2} \). When this prism is set for minimum angle of deviation, it exactly meets the condition of critical angle. The prism angle is _____.

(A) 60°

(B) 30°

(D) 45°

Correct Answer: 60°

Step-by-Step Solution

For a prism at minimum deviation, the refraction angles inside the prism are equal.

\[ r_1 = r_2 = \frac{A}{2} \]

At the exit surface, the ray meets the critical angle condition.

Critical angle \(C\) is given by:

\[ \sin C = \frac{\text{Refractive index of rarer medium}} {\text{Refractive index of denser medium}} \]

Here,

\[ \sin C = \frac{(n/2)}{n} = \frac{1}{2} \]

\[ C = 30^\circ \]

Since the ray inside the prism makes angle \( \dfrac{A}{2} \) with the normal at the exit face and it just satisfies the critical angle condition,

\[ \frac{A}{2} = C = 30^\circ \]

\[ A = 60^\circ \]

Therefore, the prism angle is \[ \boxed{60^\circ} \]

Step-by-Step Solution

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