Small Angle Prism Formula

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 $$4.8^\circ$$.

(A) $$\sin i_1 = \sin e_1 = n_1 \sin \frac{A_1}{2}$$

$$\sin i_2 = \sin e_2 = n_2 \sin \frac{A_2}{2}$$

$$e_1 = i_2 \Rightarrow n_1 \sin \frac{A_1}{2} = n_2 \sin \frac{A_2}{2}$$

(B) $$i_2 = e_2 = e_1$$

$$\sin i_2 = n_2 \sin \frac{A_2}{2}$$

But $$i_1 \neq i_2$$

(C) $$\delta_1 = (\mu_1 - 1) A_1, \delta_2 = (\mu_2 - 1) A_2$$

$$\frac{1}{2} \left( \frac{\delta_1}{\mu_1 - 1} \right) + \frac{1}{2} \left( \frac{\delta_2}{\mu_2 - 1} \right) = \frac{1}{2} \frac{A_1}{2} + \frac{A_2}{2} = \theta$$

(D) $$i_1 = e_1 = i_2$$

$$\sin i_1 = n_1 \sin \frac{A_1}{2}$$