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JEE Main 2021 (31 Aug, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2021 (31 Aug, Shift 2)

Q61JEE Main 2021 · 31 AUG · SHIFT 2Medium

Prism, Dispersion & Scattering

Cross-section view of a prism is the equilateral triangle ABC in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from P (midpoint of BC) to A is ______________ ×\times 10−-10 s. (Given, speed of light in vacuum = 3 ×\times 108 m/s and cos30∘=32{}^\circ = \frac{\sqrt{3}}{2})

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Numerical answer type. Work it out, then open the answer.

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Answer: 5

  1. Minimum deviation with i=A=60∘i = A = 60^\circ and r=A2=30∘r = \frac{A}{2} = 30^\circ: μ=sin⁡60∘sin⁡30∘=3\mu = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \sqrt{3}.
  2. At minimum deviation the ray inside is parallel to the base BC; distance PA=10cos⁡30∘=53 cmPA = 10\cos 30^\circ = 5\sqrt{3}\text{ cm} (altitude of the triangle).
  3. Speed in prism v=cμ=3×1083 m/sv = \frac{c}{\mu} = \frac{3\times10^8}{\sqrt{3}}\text{ m/s}
  4. t=PAv=53×10−2×33×108=5×10−10 st = \dfrac{PA}{v} = \dfrac{5\sqrt{3}\times10^{-2}\times\sqrt{3}}{3\times10^8} = 5\times10^{-10}\text{ s}
  5. So the answer is 5.
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