If a thin prism of glass is dipped into water then minimum deviation (with respect to air) of light produced by prism will be: $\left({}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{g}}=\frac{3}{2} \mathrm{and} {}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{w}}=\frac{4}{3}\right)$  

The angle of minimum deviation in thin prism is given by:

$$\begin{gathered} \mathrm{\delta } = (\mathrm{\mu } - 1)\mathrm{A}, \\ \mathrm{where} \mathrm{\mu } \mathrm{is} \mathrm{refractive} \mathrm{index} \mathrm{of} \mathrm{prism} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{surrounding} \mathrm{medium}. \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{\delta }}_{\mathrm{a}}}{{\mathrm{\delta }}_{\mathrm{w}}}=\frac{\left({}^{\mathrm{a}}{\mathrm{\mu }}_{\mathrm{g}}-1\right)A}{\left({}^{\mathrm{w}}{\mathrm{\mu }}_{\mathrm{g}}-1\right)A}=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3/2}{4/3}-1\right)}=4 \\ \Rightarrow {\mathrm{\delta }}_{\mathrm{w}}=\frac{{\mathrm{\delta }}_{\mathrm{a}}}{4} \end{gathered}$$