A thin glass prism of refractive index $1.5$ produces a deviation of $4°$ of a ray incident at a small angle. What will be the deviation of the same incident ray by the same prism if it is immersed in water of refractive index $\frac{4}{3}$?

For a small-angled prism and for a small angle of incidence, the deviation is given by
$\delta =({\mu }_g-1) A$ ….. (i)
The refractive index of the prism, when it is dipped in water is
$$\begin{gathered} {\mu }^{\text{'}}=\frac{{\mu }_g}{{\mu }_w} \\ =\frac{1.5}{{\displaystyle \frac{4}{3}}} \\ =\frac{9}{8} \end{gathered}$$

${\delta }^{\text{'}}=({\mu }^{\text{'}}-1) A$….. (ii)

From (i) and (ii) we have
$$\begin{gathered} {\delta }^{\text{'}}=\left(\frac{{\mu }^{\text{'}}-1}{{\mu }_g-1}\right)\delta \\ =\frac{\left({\displaystyle \frac{9}{8}}-1\right)}{(1.5-1)}\times 4^o \\ =1^o \end{gathered}$$