Skip to content

Question

A prism of refracting angle \( 30^\circ \) and refractive index \( 1.6 \) produces what minimum
deviation?

Options

Choose one · Correct answer highlighted

Explanation

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. For a thin prism: D_m = (n - 1) A . n = 1.6 , A = 30° . D_m = (1.6 - 1) × 30 = 0.6 × 30 = 18° . Substituting values gives 18°, 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.

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.