The refractive index of a transparent liquid filled in an...

For minimum deviation in a prism,

\mu = \dfrac{\sin \left(\dfrac{{D}_\min + A}{2}\right)}{sin\dfrac{A}{2}}

The value of minimum deviation is

⇒ \sqrt{2} = \dfrac{\sin \left(\dfrac{{D}_\min + 60 °}{2}\right)}{\sin 30 °}

⇒ \dfrac{1}{\sqrt{2}} = \sin \left(\dfrac{{D}_\min + 60 °}{2}\right) ⇒ 45 ° = \left(\dfrac{{D}_\min + 60 °}{2}\right) ⇒ {D}_\min = 30 °