For an achromatic combination of two thin prisms

If the prisms are of crown glass and flint glass, then for achromatism: $\omega_c \delta_c = \omega_f \delta_f$ where $\omega$ is dispersive power and $\delta$ is deviation for mean (yellow) ray.

Deviation produced by a thin prism

For yellow light, $\delta = (\mu - 1)A$ where $\mu$ is refractive index for yellow light and $A$ is prism angle.

So, $\delta_c = (\mu_c-1)A_c = (1.5-1)A_c = 0.5A_c$ $\delta_f = (\mu_f-1)A_f = (1.6-1)A_f = 0.6A_f$

Apply achromatic condition

$\omega_c\delta_c = \omega_f\delta_f$ $0.02\,(0.5A_c)=0.03\,(0.6A_f)$ $0.01A_c=0.018A_f$ $A_f=\frac{0.01}{0.018}A_c=\frac{5}{9}A_c$

Net deviation is given as $2^\circ$

In an achromatic combination, the two prisms are oppositely placed, so net deviation is: $\delta_{\text{net}}=\delta_c-\delta_f=2^\circ$

Substitute: $0.5A_c-0.6A_f=2$

Using $A_f=\frac{5}{9}A_c$, $0.5A_c-0.6\left(\frac{5}{9}A_c\right)=2$ $0.5A_c-\frac{3}{10}\cdot\frac{5}{?}$

Compute carefully: $0.6\cdot\frac{5}{9}=\frac{3}{5}\cdot\frac{5}{9}=\frac{1}{3}$ Hence, $0.5A_c-\frac{1}{3}A_c=2$ $\left(\frac{1}{2}-\frac{1}{3}\right)A_c=2$ $\frac{1}{6}A_c=2$ $A_c=12^\circ$

Rounded to nearest integer

$A_c = 12^\circ$

Comparison with stored answer

Stored correct answer = $12$

Our derived answer matches the stored answer.