A beam of polychromatic light passes through a thin prism of prism...

For a thin prism, the angle of deviation is given by $D = (n - 1)A$.

To find the smallest value of deviation $D_m$, we need to find the minimum value of the refractive index $n(\lambda)$.

Given $n(\lambda) = \alpha\lambda + \dfrac{\beta}{\lambda^2}$.

Differentiating $n(\lambda)$ with respect to $\lambda$ and equating to zero for minimum:

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

$\lambda^3 = \dfrac{2\beta}{\alpha}$

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

$\lambda^3 = \dfrac{2 \times 0.096}{3} = 0.064$

$\lambda = 0.4\ \mu\text{m}$

Now, substituting $\lambda = 0.4\ \mu\text{m}$ back into the expression for $n(\lambda)$:

n_\min = 3(0.4) + \dfrac{0.096}{(0.4)^2}

n_\min = 1.2 + \dfrac{0.096}{0.16} = 1.2 + 0.6 = 1.8

The smallest angle of deviation is:

D_m = (n_\min - 1)A

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

Answer: $4.8^\circ$