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MHT CET 2022 (11 Aug, Shift 2) · Total Internal Reflection

Total Internal Reflection question from MHT CET 2022 (11 Aug, Shift 2)

Q23MHT CET 2022 · 11 AUG · SHIFT 2Easy

Total Internal Reflection

The speed of light in two media M1M_1 and M2M_2 are 1.5×108 m/s1.5 \times 10^{8}\text{ m/s} and 2×108 m/s2 \times 10^{8}\text{ m/s} respectively. If the light undergoes total internal reflection, the critical angle between the two media is

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Answer: (d)

  1. TIR occurs from the slower (denser) medium M1M_1 to the faster medium M2M_2.
  2. sin⁡C=μ2μ1=v1v2=1.5×1082×108=34\sin C = \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2} = \frac{1.5 \times 10^8}{2 \times 10^8} = \frac{3}{4}
  3. So the answer is (d) sin⁡−1(34)\sin^{-1}\left(\frac{3}{4}\right).
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