JEE Main 2026 Apr 2 Shift 1, Physics Q15: Prism

JEE Main2026Apr 2, Shift 1Physics

For a thin symmetric prism made of glass (refractive index $1.5$ ), the ratio of incident angle and minimum deviation will be _____.

Solution

For a thin prism, the minimum deviation is $δ_{m} = (μ - 1) A = (1.5 - 1) A = \frac{A}{2}$ .

At minimum deviation, $i = \frac{A + δ _{m}}{2} = \frac{A + A /2}{2} = \frac{3 A}{4}$ .

$∴ \frac{i}{δ _{m}} = \frac{3 A /4}{A /2} = \frac{3}{2}$

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