When a glass prism is immersed in water, its dispersive power

The dispersive power of a glass prism in air is given by:

$\omega = \dfrac{\mu_v - \mu_r}{\mu_y - 1}$

where $\mu_v$, $\mu_r$, and $\mu_y$ are the refractive indices of glass for violet, red, and yellow (mean) light, respectively.

When the prism is immersed in water, the relative refractive index of glass with respect to water is considered. Assuming the refractive index of water $\mu_w$ is approximately constant for all colors (ignoring the dispersion of water), the relative refractive indices are:

$\mu'_v = \dfrac{\mu_v}{\mu_w}$, $\mu'_r = \dfrac{\mu_r}{\mu_w}$, and $\mu'_y = \dfrac{\mu_y}{\mu_w}$

The new dispersive power $\omega'$ in water becomes:

$\omega' = \dfrac{\mu'_v - \mu'_r}{\mu'_y - 1}$

Substituting the relative refractive indices, we get:

$\omega' = \dfrac{\dfrac{\mu_v}{\mu_w} - \dfrac{\mu_r}{\mu_w}}{\dfrac{\mu_y}{\mu_w} - 1} = \dfrac{\mu_v - \mu_r}{\mu_y - \mu_w}$

Comparing the denominators of $\omega$ and $\omega'$:

Since the refractive index of water $\mu_w > 1$, it follows that $\mu_y - \mu_w < \mu_y - 1$.

Because the denominator decreases while the numerator remains unchanged, the overall value of the fraction increases.

Therefore, $\omega' > \omega$, meaning the dispersive power of the prism increases when immersed in water.

Answer: increases