Question

Sol. [Key Concept: $\delta_{m} = (\mu - 1)A$ and $i = (A + \delta_{m})/2$; form the ratio $i/\delta_{m}$.]

We have for thin prism minimum deviation :

$${\delta_{m} = (\mu - 1)A }{\Rightarrow \delta_{m} = (1.5 - 1)A\ \Rightarrow \delta_{m} = \frac{A}{2} }$$Also, $i = \frac{A + \delta_{m}}{2} \Rightarrow i = \frac{A + \frac{A}{2}}{2} \Rightarrow i = \frac{3A}{4}$

Thus $\frac{i}{\delta_{m}} = \frac{3}{2}$