Minimum Angle of Deviation for Prism with Wavelength Dependent Refractive Index

For a thin prism of angle $A$, the angle of minimum deviation $D_m$ for light of wavelength $\lambda$ is given by: $D_m(\lambda) = (n(\lambda) - 1) A$

Here, the prism angle is given as $A = 6^\circ$. The refractive index as a function of wavelength $\lambda$ is: $n(\lambda) = \alpha \lambda + \frac{\beta}{\lambda^2}$

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

Since $A$ is constant and positive, the angle of minimum deviation $D_m(\lambda)$ is minimized when the refractive index $n(\lambda)$ is minimized.

To find the wavelength $\lambda_{\text{min}}$ at which $n(\lambda)$ reaches its minimum, we differentiate $n(\lambda)$ with respect to $\lambda$ and set the derivative to zero: $\frac{dn}{d\lambda} = \alpha - \frac{2\beta}{\lambda^3} = 0$

Solving for $\lambda$: $\lambda^3 = \frac{2\beta}{\alpha}$

Substitute the given values of $\alpha$ and $\beta$: $\lambda_{\text{min}}^3 = \frac{2 \times 0.096\ \mu\text{m}^2}{3\ \mu\text{m}^{-1}} = \frac{0.192}{3}\ \mu\text{m}^3 = 0.064\ \mu\text{m}^3$

Taking the cube root on both sides gives: $\lambda_{\text{min}} = (0.064)^{1/3}\ \mu\text{m} = 0.4\ \mu\text{m}$

To confirm that this represents a minimum, we evaluate the second derivative: $\frac{d^2n}{d\lambda^2} = \frac{6\beta}{\lambda^4} > 0 \quad \text{for } \lambda > 0$ Since the second derivative is positive, $n(\lambda)$ is indeed minimized at $\lambda_{\text{min}} = 0.4\ \mu\text{m}$.

Now, calculating the refractive index at $\lambda_{\text{min}}$: $n(\lambda_{\text{min}}) = \alpha \lambda_{\text{min}} + \frac{\beta}{\lambda_{\text{min}}^2}$ $n(\lambda_{\text{min}}) = (3)(0.4) + \frac{0.096}{(0.4)^2} = 1.2 + \frac{0.096}{0.16} = 1.2 + 0.6 = 1.8$

Finally, substituting $n(\lambda_{\text{min}}) = 1.8$ into the deviation formula: $D_m = (1.8 - 1) \times 6^\circ = 0.8 \times 6^\circ = 4.8^\circ$

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

Correct Answer: B ($4.8^\circ$)