The ratio of the angle of deviation produced by a thin prism, when it is placed in air to the angle of deviation produced when it is immersed in water of refractive index $\dfrac{4}{3}$ is:

The angle of deviation produced by a thin prism is given by $\delta = (\mu_{rel} - 1)A$, where $\mu_{rel}$ is the refractive index of the prism relative to the surrounding medium.

Assuming the refractive index of the glass prism to be $\mu_g = \dfrac{3}{2}$.

When the prism is in air, the relative refractive index is $\mu_g$.

$\delta_{air} = \left(\mu_g - 1\right)A = \left(\dfrac{3}{2} - 1\right)A = \dfrac{A}{2}$

When the prism is immersed in water, the relative refractive index is $\mu_{rel} = \dfrac{\mu_g}{\mu_w}$.

Given $\mu_w = \dfrac{4}{3}$, we have $\mu_{rel} = \dfrac{3/2}{4/3} = \dfrac{9}{8}$.

$\delta_{water} = \left(\dfrac{9}{8} - 1\right)A = \dfrac{A}{8}$

The ratio of the angle of deviation in air to that in water is:

$\dfrac{\delta_{air}}{\delta_{water}} = \dfrac{\dfrac{A}{2}}{\dfrac{A}{8}} = 4$

Thus, the ratio is $4:1$.

Answer: $4:1$