Question

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index \(n\) of the prism satisfies:

• A. \(\sqrt{2}<n<2\)
• B. \(\sqrt{2}<n<2\sqrt{2}\)
• C. \(n\ge 2\)
• D. \(1<n<2\)

Hint:

When minimum deviation equals prism angle, always substitute \(\delta=A\) directly into the prism formula before simplifying.

Answer 1

The Correct Option is A

Concept: For a prism, the relation between refractive index \(n\), angle of prism \(A\), and angle of minimum deviation \(\delta\) is: \[ n=\frac{\sin\left(\frac{A+\delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Step 1: Use the given condition Given: \[ \delta=A \] Substitute in the formula: \[ n=\frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} =\frac{\sin A}{\sin\left(\frac{A}{2}\right)} \] Using \(\sin A=2\sin\frac{A}{2}\cos\frac{A}{2}\), \[ n=2\cos\frac{A}{2} \]
Step 2: Determine the range of \(n\) For a prism: \[ 0<A<\pi \Rightarrow 0<\frac{A}{2}<\frac{\pi}{2} \] Hence: \[ 0<\cos\frac{A}{2}<1 \] Thus: \[ 0<n<2 \] For a real transparent prism, the minimum deviation condition requires: \[ A>60^\circ \Rightarrow \cos\frac{A}{2}<\cos30^\circ=\frac{\sqrt3}{2} \] Hence: \[ n=2\cos\frac{A}{2}>\sqrt2 \] Final Answer: \[ \boxed{\sqrt{2}<n<2} \]