Question 3

SCQHARD

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 $n(\lambda) = a\lambda + \frac{b}{\lambda^2}$ , where $a = 3\text{ }\mu\text{m}^{-1}$ and $b = 0.096\text{ }\mu\text{m}^2$ . 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

(A)

$6.4^\circ$

(B)

$4.8^\circ$

(C)

$3.2^\circ$

(D)

$2.4^\circ$

Detailed Solution

Relation for thin prism deviation: For a thin prism, $D_m = (n - 1)A$ . $D_m$ is smallest when $n(\lambda)$ is minimum.
Minimize $n(\lambda)$ : $n(\lambda) = a\lambda + b\lambda^{-2}$ $\frac{dn}{d\lambda} = a - \frac{2b}{\lambda^3} = 0 \implies \lambda^3 = \frac{2b}{a}$ Given $a = 3$ , $b = 0.096$ : $\lambda^3 = \frac{2 \times 0.096}{3} = 0.064$ $\lambda = \sqrt[3]{0.064} = 0.4\text{ }\mu\text{m}$ .
Calculate minimum refractive index ( $n_{min}$ ): $n(0.4) = 3(0.4) + \frac{0.096}{(0.4)^2} = 1.2 + \frac{0.096}{0.16} = 1.2 + 0.6 = 1.8$ .
Calculate smallest $D_m$ : $D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$ .

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