Geometrical Optics

Concept:For a thin prism, the angle of minimum deviation is $D_m = (n - 1)A$, where $A$ is the prism angle and $n$ is the refractive index. To find the smallest $D_m$, we must find the wavelength $\lambda$ at which $n$ is minimum.Explanation:Step 1: Write the refractive index function: $n(\lambda) = \alpha \lambda + \frac{\beta}{\lambda^2}$, with $\alpha = 3\ \mu\text{m}^{-1}$ and $\beta = 0.096\ \mu\text{m}^2$.Step 2: Differentiate $n$ with respect to $\lambda$: $\frac{dn}{d\lambda} = \alpha - \frac{2\beta}{\lambda^3}$.Step 3: Set $\frac{dn}{d\lambda} = 0$ to find critical points: $\alpha - \frac{2\beta}{\lambda_{\min}^3} = 0$.Step 4: Solve for $\lambda_{\min}^3$: $\lambda_{\min}^3 = \frac{2\beta}{\alpha} = \frac{2 \times 0.096}{3} = 0.064\ \mu\text{m}^3$.Step 5: Take cube root: $\lambda_{\min} = (0.064)^{1/3} = 0.4\ \mu\text{m}$.Step 6: Verify minimum using second derivative: $\frac{d^2 n}{d\lambda^2} = \frac{6\beta}{\lambda^4} > 0$ (since $\beta > 0$), confirming a minimum.Step 7: Calculate $n$ at $\lambda_{\min}$: $n = \alpha \lambda_{\min} + \frac{\beta}{\lambda_{\min}^2} = 3 \times 0.4 + \frac{0.096}{(0.4)^2} = 1.2 + \frac{0.096}{0.16} = 1.2 + 0.6 = 1.8$.Step 8: Use thin prism formula: $D_m = (n - 1)A = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$.Thus, the minimum deviation at $\lambda_{\min}$ is $4.8^\circ$.Answer:$4.8^\circ$ (Option B)