The refractive index of the material of the prism is √(3), then the angle of minimum deviation of the prism is (a) 30° (b) 45° (c) 60° (d) 75°

Answer: (c). Take the prism as equilateral, A = 60°. μ = (sin (A+δm)/(2))/(/2) ⇒ sin (60°+δm)/(2) = √(3) × 1/2 = √(3)/2 (60°+δm)/(2) = 60° ⇒ δm = 60° So the answer is (c) 60°.

The refractive index of the material of the prism is √(3), then the… | NEET 1999 (AIPMT)

Q91NEET 1999 · AIPMTEasy

Prism, Dispersion & Scattering

The refractive index of the material of the prism is $\sqrt{3}$ , then the angle of minimum deviation of the prism is

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Answer: (c)

Take the prism as equilateral, $A = 60^\circ$ .
$\mu = \dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac{A}{2}} \Rightarrow \sin\frac{60^\circ+\delta_m}{2} = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}$
$\frac{60^\circ+\delta_m}{2} = 60^\circ \Rightarrow \delta_m = 60^\circ$
So the answer is (c) $60^\circ$ .

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