A prism is placed in water. The angle of minimum deviation

When a prism is placed in water then we get

Refractive index of prism material with respect to water as

${ }_w{\mu }_g^{|}=\frac{\sin \left(\frac{\text{A}+{\text{d}}_{\text{m}}^{\text{|}}}{2}\right)}{\sin \text{A/2}}$ 

${\text{d}}_{\text{m}}^{\text{|}}$is minimum deviation produced

Let us consider the refractive index of prism material w.r.to air

${ }_{air}{\mu }_g=\frac{\sin (\text{A}+{\text{d}}_{\text{m}})}{\sin \text{A/2}}$

We get ${ }_{air}{\mu }_g{>}_{water}{\mu }_g^{|}$            $({\because }_{\text{air}}{\mu }_{\text{g}}={\text{3/2,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}_{\omega }{\mu }_{\text{g}}=\text{9/8})$

$\Rightarrow \left(\text{air}\right){\text{d}}_{\text{m}}>{\text{d}}_{\text{m}}^{\text{|}}\left(\text{in\hspace{0.17em}water}\right)$

The minimum deviation will decrease in the case of prism immersed in the

water when compared with minimum deviation produced when prism is

kept in air.