A beam of polychromatic light passes through a thin prism of prism angle 6°. The refractive index of the material of the prism varies with wavelength (λ) as $n\left(\lambda \right)=a\lambda +\frac{b}{{\lambda }^2}$, where a = 3 µm⁻¹ and b = 0.096 µm². If λ_min is the wavelength at which the angle of minimum deviation D_m is smallest, then the correct value of D_m at λ_min is

Correct Answer: The correct option is 4.8°.

Step-by-Step Explanation:

1. Deviation produced by a thin prism:

For a thin prism of prism angle $A$ and refractive index $n$, the angle of minimum deviation $D_m$ is given by the formula:

$D_m=(n-1)A$

Here, the prism angle is given as $A=6°$.

2. Finding the wavelength ${\lambda }_{\text{min}}$ for smallest deviation:

Since $A$ is a constant, $D_m$ is minimum when the refractive index $n\left(\lambda \right)$ is minimum.

The refractive index varies with wavelength as:

$n\left(\lambda \right)=a\lambda +\frac{b}{{\lambda }^2}$

To minimize $n\left(\lambda \right)$, we take the first derivative with respect to $\lambda$ and set it equal to zero:

$\frac{dn}{d\lambda }=a-\frac{2b}{{\lambda }^3}=0$

Solving for $\lambda$:

${\lambda }^3=\frac{2b}{a}$

Substituting the given values $a=3\text{ µm}{⁻}^1$ and $b=0.096\text{ µm}{²}^{\mathrm{}}$:

${\lambda }^3=\frac{2\times 0.096}{3}=\frac{0.192}{3}=0.064\text{ µm}{³}^{\mathrm{}}$

Taking the cube root on both sides:

${\lambda }_{\text{min}}=\sqrt[3]{0.064}=0.4\text{ µm}$

3. Calculating the minimum refractive index:

Now, substitute ${\lambda }_{\text{min}}=0.4\text{ µm}$ back into the refractive index formula:

$n_{\text{min}}=a{\lambda }_{\text{min}}+\frac{b}{{\lambda }_{\text{min}}^2}$

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

$n_{\text{min}}=1.2+\frac{0.096}{0.16}=1.2+0.6=1.8$

4. Calculating the smallest angle of minimum deviation $D_m$:

Now substitute $n_{\text{min}}=1.8$ and $A=6°$ into the deviation equation:

$D_m=(1.8-1)\times 6°$

$D_m=0.8\times 6°=4.8°$

Thus, the correct value of $D_m$ at ${\lambda }_{\text{min}}$ is 4.8°.