JEE Advanced 2026 · Paper 2 · Physics Q3

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A beam of polychromatic light passes through a thin prism of prism angle $6^\circ$ . The refractive index of the material of the prism varies with wavelength $(\lambda)$ as $n(\lambda)=\alpha\lambda+\dfrac{\beta}{\lambda^2}$ , where $\alpha=3\ \mu\text{m}^{-1}$ and $\beta=0.096\ \mu\text{m}^2$ . If $\lambda_{\min}$ is the wavelength at which the angle of minimum deviation $D_m$ is smallest, then the correct value of $D_m$ at $\lambda_{\min}$ is

6.4^\circ

4.8^\circ

3.2^\circ

2.4^\circ

Answer and worked solution

Official answer: B

For a thin prism $D_m=(n-1)A$ , so $D_m$ is smallest where $n(\lambda)$ is smallest:
$\frac{dn}{d\lambda}=\alpha-\frac{2\beta}{\lambda^3}=0\implies\lambda^3=\frac{2\times0.096}{3}=0.064\implies\lambda_{\min}=0.4\ \mu\text{m}.$
Then $n=3(0.4)+\dfrac{0.096}{0.16}=1.2+0.6=1.8$ , and
$D_m=(1.8-1)\times6^\circ=4.8^\circ.$

Topics:Prism and dispersion

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