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Question

A prism of angle \( 30^\circ \) has a minimum deviation of \( 20^\circ \). What is the refractive
index?

Options

Choose one · Correct answer highlighted

Explanation

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. Refractive index: n = (sin ( (A + D_m/2) )/sin ( (A/2) )) . A = 30° , D_m = 20° . n = (sin ( (30 + 20/2) )/sin ( (30/2) )) = (sin 25°/sin 15°) . sin 25° ≈ 0.423 , sin 15° ≈ 0.259 . n = (0.423/0.259) ≈ 1.63 . Substituting values gives 1.63, which matches expected image position and magnification from mirror/lens formula 1/f =

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