A ray enters a spherical liquid drop from air at angle of incidence θ ≠ 0. The drop has refractive index μ = √3. It undergoes exactly one internal reflection (not total internal reflection) and emerges. For minimum total deviation, determine 12 sin²θ. [The source printed key is 9; independent optics gives 4.]
Physics · Mixed Chapter
JEE Main Practice

Question

A ray enters a spherical liquid drop from air at angle of incidence θ ≠ 0. The drop has refractive index μ = √3. It undergoes exactly one internal reflection (not total internal reflection) and emerges. For minimum total deviation, determine 12 sin²θ. [The source printed key is 9; independent optics gives 4.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — For one internal reflection, total deviation is δ = 180° + 2i − 4r, and Snell law is sin i = μ sin r. Step 2 — At minimum deviation dδ/di = 0, giving dr/di = 1/2. Differentiating Snell law: cos i = μ cos r(dr/di), so 2 cos i = μ cos r. Step 3 — Let s = sin²i. Squaring Snell gives sin²r = s/μ². Squaring the minimum-deviation relation gives 4(1−s) = μ²(1−s/μ²) = μ²−s. Step 4 — Rearranging 3s = 4−μ² ⇒ s=(4−μ²)/3. With μ=√3, μ²=3, s=1/3. Step 5 — 12 sin²θ = 12×1/3 = 4. QUALITY FLAG — Origi…