If the refractive index of the material of a prism is cot (A/2), where A is the angle of prism then the angle of minimum deviation will be (a) π - 2A (b) π/2 - 2A (c) π - A (d) π/2 - A

Answer: (a). μ = (sin (A+δm)/(2))/(/2) = /2 = (/2)/(/2) So sin (A+δm)/(2) = /2 = sin (π/2 - A/2) (A+δm)/(2) = π/2 - A/2 ⇒ δm = π - 2A So the answer is (a) π - 2A.

If the refractive index of the material of a prism is cot (A/2)… | JEE Main 2024 (27 Jan, Shift 1)

Q35JEE Main 2024 · 27 JAN · SHIFT 1Medium

Prism, Dispersion & Scattering

If the refractive index of the material of a prism is $\cot\left(\frac{A}{2}\right)$ , where $A$ is the angle of prism then the angle of minimum deviation will be

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Answer: (a)

$\mu = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} = \cot\frac{A}{2} = \dfrac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$
So $\sin\frac{A+\delta_m}{2} = \cos\frac{A}{2} = \sin\left(\frac{\pi}{2} - \frac{A}{2}\right)$
$\frac{A+\delta_m}{2} = \frac{\pi}{2} - \frac{A}{2} \Rightarrow \delta_m = \pi - 2A$
So the answer is (a) $\pi - 2A$ .

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