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

The question involves finding the range of the refractive index (\(\mu\)) of the material of a prism when the minimum deviation produced by the prism is equal to its refracting angle. Let's solve this step-by-step:

1. In the condition of minimum deviation (\(D_m\)), the angle of deviation is given by: D_m = 2i - A, where \(A\) is the refracting angle of the prism and \(i\) is the angle of incidence equal to the angle of emergence.
2. For the minimum deviation to be equal to the refracting angle of the prism, we have: D_m = A. Therefore, substituting, we get: A = 2i - A \Rightarrow i = A.
3. Using Snell's Law at the point of incidence, we get: \mu = \frac{\sin i}{\sin r} = \frac{\sin A}{\sin \frac{A}{2}}.
4. Under the given condition, \mu = \frac{\sin A}{\sin \frac{A}{2}}, with \(A = D_m = 2 \times r\), so we have: \sin A = 2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}.
5. Thus: \mu = \frac{2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}}{\sin \frac{A}{2}} = 2 \times \cos \frac{A}{2}.
6. Since \(\cos \frac{A}{2}\) ranges from 0 to 1, the value of \(\mu\) will range between 1 and \(\sqrt{2}\) as \(\cos \frac{A}{2} = \frac{1}{\sqrt{2}}\).

Therefore, the correct range for the refractive index of the prism material is: 1 < \mu < \sqrt{2}.

This matches with the given correct option.