Light gets deviated by 60° from an equilateral prism for two angles of incidence differing by 30°. The angle (s) of incidence is (are): 

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angle} \mathrm{of} \mathrm{incidence} \mathrm{be} \mathrm{i} \mathrm{and} \mathrm{i}+{30}^0 \\ \mathrm{Given} \mathrm{that} \mathrm{the} \mathrm{angle} \mathrm{of} \mathrm{deviation} \mathrm{is} \mathrm{same}, \\ \mathrm{It} \mathrm{is} \mathrm{possible} \mathrm{only} \mathrm{when} \mathrm{the} \mathrm{emergent} \mathrm{ray} \mathrm{acts} \mathrm{as} \mathrm{the} \mathrm{incident} \mathrm{ray}. \\ \mathrm{So} \mathrm{angle} \mathrm{of} \mathrm{emergence} \mathrm{is} \mathrm{i}+{30}^0 , \mathrm{when} \mathrm{the} \mathrm{angle} \mathrm{of} \mathrm{incidence} \mathrm{is} \mathrm{i}. \mathrm{vice} \mathrm{versa} \\ \begin{array}{l}\delta =i+e-A\\ {60}^{\circ }=i+i+{30}^{\circ }-{60}^{\circ }\Rightarrow i={45}^{\circ }\text{ and }i+{30}^{\circ }={75}^{\circ }\end{array} \end{gathered}$$