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JEE Main 2022 (27 Jul, Shift 2) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2022 (27 Jul, Shift 2)

Q54JEE Main 2022 · 27 JUL · SHIFT 2Medium

Prism, Dispersion & Scattering

A thin prism of angle 6∘6^\circ and refractive index for yellow light (nY) 1.5(n_Y)\,1.5 is combined with another prism of angle 5∘5^\circ and nY=1.55n_Y = 1.55. The combination produces no dispersion. The net average deviation (δ)(\delta) produced by the combination is (1x)∘\left(\frac{1}{x}\right)^\circ. The value of xx is ____________.

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Answer: 4

  1. The prisms are inverted relative to each other, so their deviations subtract.
  2. δ=(n1−1)A1−(n2−1)A2\delta = (n_1 - 1)A_1 - (n_2 - 1)A_2
  3. δ=0.5×6∘−0.55×5∘=3∘−2.75∘=0.25∘\delta = 0.5 \times 6^\circ - 0.55 \times 5^\circ = 3^\circ - 2.75^\circ = 0.25^\circ
  4. δ=(14)∘⇒x=4\delta = \left(\tfrac{1}{4}\right)^\circ \Rightarrow x = 4
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