Ray Optics

JEE Main level, single correct, medium

236 of 295

Question 236 of 295

For a transparent prism, if the angle of minimum deviation is equal to its refracting angle, the refractive index nn of the prism satisfies.

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Answer: (C) 2<n<2\sqrt{2} \lt n \lt 2

At minimum deviation, δm=A\delta_m = A (angle of minimum deviation equals prism angle). The refractive index formula becomes:
n=sin⁡(A+δm2)sin⁡(A2)=sin⁡(2A2)sin⁡(A2)=sin⁡Asin⁡(A2)n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{2A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}
Using sin⁡A=2sin⁡(A2)cos⁡(A2)\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right):
n=2sin⁡(A2)cos⁡(A2)sin⁡(A2)=2cos⁡(A2)n = \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)
For a physical prism, 0<A<90°0 \lt A \lt 90°, so 45°<A2<90°45° \lt \frac{A}{2} \lt 90°, giving 0<cos⁡(A2)<120 \lt \cos\left(\frac{A}{2}\right) \lt \frac{1}{\sqrt{2}}.
Therefore, 1<n<2⋅2=221 \lt n \lt \sqrt{2} \cdot 2 = 2\sqrt{2}. However, more precisely, 2<n<2\sqrt{2} \lt n \lt 2.

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