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MHT CET 2026 (19 May, Shift 1) · Total Internal Reflection

Total Internal Reflection question from MHT CET 2026 (19 May, Shift 1)

Q4MHT CET 2026 · 19 MAY · SHIFT 1Medium

Total Internal Reflection

A ray of light takes time 'tt' to travel a distance 'dd' in air and it travels a distance '4d4d', in time 'TT' in a certain medium. If T=8tT = 8t, the critical angle for the pair of media is

[sin⁡(π6)=0.5,sin⁡(π4)=12,sin⁡(π3)=32,sin⁡(π2)=1]\left[\sin\left(\frac{\pi}{6}\right) = 0.5, \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \sin\left(\frac{\pi}{2}\right) = 1\right]

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Answer: (d)

  1. Speed in air c=dtc = \frac{d}{t}; speed in medium v=4dT=4d8t=d2tv = \frac{4d}{T} = \frac{4d}{8t} = \frac{d}{2t}.
  2. n=cv=2n = \frac{c}{v} = 2.
  3. sin⁡C=1n=12⇒C=π6\sin C = \frac{1}{n} = \frac{1}{2} \Rightarrow C = \frac{\pi}{6}.
  4. So the answer is (d) πc6\frac{\pi^c}{6}.
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