Question 3

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 $D_m$: $D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$.