A student sets a glass prism of angle 60.0° at the position of minimum deviation for sodium light…

Step 1. The angle of incidence. $i = \dfrac{A + \delta_m}{2} = \dfrac{60.0^\circ + 40.0^\circ}{2} = 50.0^\circ$ and $r_1 = \dfrac{A}{2} = 30.0^\circ$.

Step 2. The index. $n = \dfrac{\sin 50.0^\circ}{\sin 30.0^\circ} = \dfrac{0.7660}{0.5000} = 1.532$, so $n = 1.53$.

Step 3. The new ray. For $i = 40.0^\circ$: $\sin r_1 = \dfrac{\sin 40.0^\circ}{1.532} = \dfrac{0.6428}{1.532} = 0.4196$, so $r_1 = 24.8^\circ$. Then $r_2 = 60.0^\circ - 24.8^\circ = 35.2^\circ$ and $\sin e = 1.532 \times \sin 35.2^\circ = 0.8830$, so $e = 62.0^\circ$.

Step 4. The deviation. $\delta = i + e - A = 40.0^\circ + 62.0^\circ - 60.0^\circ = 42.0^\circ$.