Practice question
Question
A prism of angle \( 60^\circ \) has a refractive index of \( 1.4 \). What is the angle of minimum
deviation?
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. For a thin prism: D_m = (n - 1) A . n = 1.4 , A = 60° . D_m = (1.4 - 1) × 60 = 0.4 × 60 = 24° . Substituting values gives 24°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
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