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JEE Main 2022 (28 Jun, Shift 1) · Prism, Dispersion & Scattering

Prism, Dispersion & Scattering question from JEE Main 2022 (28 Jun, Shift 1)

Q53JEE Main 2022 · 28 JUN · SHIFT 1Medium

Prism, Dispersion & Scattering

The refracting angle of a prism is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be -

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Answer: (a)

  1. sin⁡A+δm2sin⁡A2=cot⁡A2⇒sin⁡A+δm2=cos⁡A2=sin⁡(90∘−A2)\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} = \cot\frac{A}{2} \Rightarrow \sin\frac{A+\delta_m}{2} = \cos\frac{A}{2} = \sin\left(90^\circ - \frac{A}{2}\right)
  2. A+δm2=90∘−A2⇒δm=180∘−2A\frac{A+\delta_m}{2} = 90^\circ - \frac{A}{2} \Rightarrow \delta_m = 180^\circ - 2A
  3. So the answer is (a) 180∘−2A180^\circ - 2A.
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