The angle of minimum deviation produced by a thin prism in air is...

For a thin prism, the angle of minimum deviation is given by $\delta = (n - 1)A$, where $n$ is the relative refractive index of the prism with respect to the surrounding medium and $A$ is the angle of the prism.

When the prism is in air, the deviation is:

$\delta_1 = ({}_a n_g - 1)A = \left(\dfrac{3}{2} - 1\right)A = \dfrac{A}{2}$

When the prism is immersed in the liquid, the relative refractive index of the glass with respect to the liquid is:

${}_l n_g = \dfrac{{}_a n_g}{{}_a n_l} = \dfrac{3/2}{4/3} = \dfrac{9}{8}$

The deviation in the liquid is:

$\delta_2 = ({}_l n_g - 1)A = \left(\dfrac{9}{8} - 1\right)A = \dfrac{A}{8}$

Taking the ratio of the two deviations:

$\dfrac{\delta_2}{\delta_1} = \dfrac{\dfrac{A}{8}}{\dfrac{A}{2}} = \dfrac{1}{4}$

$\delta_2 = \dfrac{\delta_1}{4}$

Answer: $\delta_2=\dfrac{\delta_1}{4}$