A student cuts a prism of angle 70° from a glass with refractive index 1.60…

Step 1. The critical angle. $\sin C = \dfrac{1}{1.60} = 0.625$, so $C = 38.7^\circ$.

Step 2. The largest prism angle. $2C = 77.4^\circ$. Her prism of $70^\circ$ is below this, so light can leave.

Step 3. The smallest angle of incidence. At the limit $r_2 = C$, so $r_1 = A - C = 70^\circ - 38.7^\circ = 31.3^\circ$. Then $\sin i_{min} = 1.60 \times \sin 31.3^\circ = 0.832$, so $i_{min} = 56.3^\circ$.

Step 4. The largest deviation. At $i = 90^\circ$, $r_1 = C = 38.7^\circ$ and $r_2 = 31.3^\circ$. Then $\sin e = 1.60 \times \sin 31.3^\circ = 0.832$ and $e = 56.3^\circ$. The deviation is $\delta = 90^\circ + 56.3^\circ - 70^\circ = 76.3^\circ$.