Question

The minimum deviation produced by a prism is equal to the refracting angle of prism, then choose the range of refractive index (\(\mu\)) of material of prism:

• A. \(1<\mu<\sqrt{2}\)
• B. \(1<\mu<2\)
• C. \(1<\mu<2\sqrt{2}\)
• D. \(1<\mu<\sqrt{3}\)

Hint:

If the angle of minimum deviation equals the prism angle, the refractive index simplifies to \(\mu = 2\cos\left(\frac{A}{2}\right)\), making range-based questions straightforward.

Answer 1

The Correct Option is A

Concept:
For a prism of refracting angle \(A\), the refractive index \(\mu\) in terms of angle of minimum deviation \(\delta_m\) is given by: \[ \mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Step 1: Apply the Given Condition
It is given that: \[ \delta_m = A \] Substitute \(\delta_m = A\) in the formula: \[ \mu = \frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)} \]
Step 2: Simplify the Expression
Using the identity: \[ \sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right) \] \[ \Rightarrow \mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\cos\left(\frac{A}{2}\right) \]
Step 3: Determine the Range of \(\mu\)
For a real prism: \[ 0^\circ<A<90^\circ \Rightarrow 0^\circ<\frac{A}{2}<45^\circ \] Hence, \[ \cos 45^\circ<\cos\left(\frac{A}{2}\right)<1 \] \[ \Rightarrow \frac{1}{\sqrt{2}}<\cos\left(\frac{A}{2}\right)<1 \] Multiplying throughout by 2: \[ \boxed{1<\mu<\sqrt{2}} \]