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JEE Main 2020 (9 Jan, Shift 2) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2020 (9 Jan, Shift 2)

Q39JEE Main 2020 · 9 JAN · SHIFT 2Medium

Total Internal Reflection

There is a small source of light at some depth below the surface of water (refractive index = 43\frac{4}{3}) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly): [Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2π\pirh]:

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Answer: (a)

  1. Only light inside the cone of half-angle CC escapes: sin⁡C=34\sin C = \frac{3}{4}, cos⁡C=74≈0.661\cos C = \frac{\sqrt{7}}{4} \approx 0.661.
  2. On a sphere of radius rr about the source, this cone cuts a cap of height h=r(1−cos⁡C)h = r(1-\cos C).
  3. Fraction =2πrh4πr2=1−cos⁡C2=1−0.6612≈0.17= \frac{2\pi r h}{4\pi r^2} = \frac{1-\cos C}{2} = \frac{1-0.661}{2} \approx 0.17.
  4. So the answer is (a) 17%.
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