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JEE Main 2017 (8 Apr, Shift 1) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2017 (8 Apr, Shift 1)

Q46JEE Main 2017 · 8 APR · SHIFT 1Medium

Total Internal Reflection

Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC. At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90°. Angle A is given by:

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Answer: (d)

  1. Snell's law (denser to rarer): n12sin⁡A=sin⁡rn_{12}\sin A = \sin r.
  2. Reflected and refracted rays are perpendicular, so A+r=90∘A + r = 90^\circ and sin⁡r=cos⁡A\sin r = \cos A.
  3. n12sin⁡A=cos⁡A⇒tan⁡A=1n12=sin⁡θCn_{12}\sin A = \cos A \Rightarrow \tan A = \frac{1}{n_{12}} = \sin\theta_C.
  4. So the answer is (d) A=tan⁡−1(sin⁡θC)A = \tan^{-1}(\sin\theta_C).
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