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JEE Main 2026 (2 Apr, Shift 2) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2026 (2 Apr, Shift 2)

Q7JEE Main 2026 · 2 APR · SHIFT 2Medium

Total Internal Reflection

One side of an equilateral prism is painted by a transparent material of refractive index n2n_2. The refractive index of prism is 1.61.6. The minimum value of n2n_2 required for total internal reflection from painted face is ________.

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Answer: (d)

  1. The ray enters the left face normally, so it goes straight. That face makes 60∘60^\circ with the base, so the ray meets the base at 60∘60^\circ to its normal.
  2. TIR at the painted base needs sin⁡60∘≥n21.6\sin60^\circ \ge \frac{n_2}{1.6}.
  3. n2≤1.6×32=0.83=435n_2 \le 1.6\times\frac{\sqrt{3}}{2} = 0.8\sqrt{3} = \frac{4\sqrt{3}}{5}.
  4. The limiting value of n2n_2 is 435\frac{4\sqrt{3}}{5} (TIR needs n2n_2 below this, so it is really a maximum). So the answer is (d).
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