Angle of minimum deviation is equal to the angle of prism A of an equilateral glass prism. The angle of incidence at which minimum deviation will be obtained is

The correct option is A

${60}^{\circ }$

$A={\delta }_m={60}^{\circ }$
At minimum deviation $i=\left(\frac{A+{\delta }_m}{2}\right)={60}^{\circ }$

We know that,
${\delta }_{min}=2i-A\Rightarrow i=\left(\frac{A+{\delta }_{min}}{2}\right)$
Now, ${\delta }_{min}=A={60}^{\circ }$ (ABC is an equilateral $\Delta$)
$i=\frac{A+{\delta }_{min}}{2}=\frac{{60}^{\circ }+{60}^{\circ }}{2}={60}^{\circ }$
Here, $i={60}^{\circ }$