JEE Main 2026 Jan 23 Shift 1, Physics Q1: Prism

JEE Main2026Jan 23, Shift 1Physics

A thin prism with angle $5^{\circ}$ of refractive index $1.72$ is combined with another prism of refractive index $1.9$ to produce dispersion without deviation. The angle of second prism is ______.

A
$4. 5^{\circ}$
B
$6^{\circ}$
C
$4^{\circ}$
D
$5^{\circ}$

Solution

For dispersion without deviation, the net deviation produced by both prisms must be zero: $δ_{1} + δ_{2} = 0$ .

Using the thin-prism deviation formula $δ = (μ - 1) A$ :

$(μ_{1} - 1) A_{1} + (μ_{2} - 1) A_{2} = 0$

$∴ A_{2} = \frac{( μ _{1} - 1 ) A _{1}}{( μ _{2} - 1 )} = \frac{( 1.72 - 1 ) \times 5 ^{\circ}}{( 1.9 - 1 )} = \frac{0.72 \times 5}{0.9} = 4^{\circ}$

(The negative sign just indicates the prism is inverted; magnitude is what's asked.)

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JEE Main 2026 Jan 23 Shift 1, Physics Q1: Prism

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