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JEE Main 202628 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2026 · 28 January · Evening Shift · Physics · Ray Optics

For a transparent prism, if the angle of minimum deviation is equal...

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

Options

  1. A1<n<21 < n < 2
  2. B2<n<22\sqrt{2} < n < 2 \sqrt{2}
  3. C2<n<2\sqrt{2} < n < 2
  4. Dn⩾2n \geqslant 2

Correct answer

C. 2<n<2\sqrt{2} < n < 2

Step-by-step solution

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 < A < 90°, so 45°<A2<90°45° < \frac{A}{2} < 90°, giving 0<cos⁡(A2)<120 < \cos\left(\frac{A}{2}\right) < \frac{1}{\sqrt{2}}.
Therefore, 1<n<2⋅2=221 < n < \sqrt{2} \cdot 2 = 2\sqrt{2}. However, more precisely, 2<n<2\sqrt{2} < n < 2.
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