An equilateral glass prism has a refractive index $$1.6$$ in the air. Calculate the angle of minimum deviation of the prism, when kept in a medium of refractive index $$4\sqrt{\dfrac{2}{5}}$$ .
When the prism is kept in another medium we have to take the refractive index of the prism with respect to the provided medium.
$$medium ^{\mu} = \dfrac{\mu_{prism}}{\mu_{medium}} = \dfrac{\sin \left[\left(\dfrac{A+D_m}{2}\right)\right]}{\sin \left(\dfrac{A}{2}\right)}$$
$$\dfrac{1.6}{\dfrac{4\sqrt{2}}{5}} = \dfrac{\sin \left[\left(\dfrac{60^o+D_m}{2}\right)\right]}{\sin \left(\dfrac{60^o}{2}\right)}$$
$$\sqrt{2} = \dfrac{\sin \left[\left(\dfrac{60^o+D_m}{2}\right)\right]}{\dfrac{1}{2}}$$
$$\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \left(\dfrac{60^o + D_m}{2}\right)$$
$$90^o = 60^o + D_m$$
$$D_m = 30^o$$