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 $$\alpha=3\,\mu\mathrm{m^{-1}}$$ and $$\beta=0.096\,\mu\mathrm{m^2}$$. 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

For a thin prism with a small angle $$A = 6^\circ$$, the deviation $$D_m$$ is given by:

$$$D_m = (n - 1)A$$$

To find the wavelength $$\lambda_{\text{min}}$$ at which $$D_m$$ is smallest, we minimize the refractive index $$n(\lambda)$$, where:

$$n(\lambda) = \alpha \lambda + \frac{\beta}{\lambda^2}$$

Differentiating $$n(\lambda)$$ with respect to $$\lambda$$ and setting the derivative to zero:

$$\frac{dn}{d\lambda} = \alpha - \frac{2\beta}{\lambda^3} = 0$$

$$\lambda_{\text{min}}^3 = \frac{2\beta}{\alpha}$$

Substituting $$\alpha = 3\text{ }\mu\text{m}^{-1}$$ and $$\beta = 0.096\text{ }\mu\text{m}^2$$:

$$\lambda_{\text{min}}^3 = \frac{2 \times 0.096}{3} = 0.064\text{ }\mu\text{m}^3$$

$$\lambda_{\text{min}} = \sqrt[3]{0.064} = 0.4\text{ }\mu\text{m}$$

Calculating the refractive index at $$\lambda_{\text{min}}$$:

$$n_{\text{min}} = 3(0.4) + \frac{0.096}{(0.4)^2} = 1.2 + 0.6 = 1.8$$

Substituting $$n_{\text{min}}$$ back into the deviation equation:

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

The correct value of $$D_m$$ at $$\lambda_{\text{min}}$$ is $$4.8^\circ$$.