A container holds water up to a height of 20 cm and a point source...

For the shadow of the ring to be formed on the ceiling, the light rays from the point source passing just outside the ring must refract into the air and reach the ceiling.

This means the angle of incidence at the water surface must be less than or equal to the critical angle $\theta_c$. If the angle of incidence exceeds $\theta_c$, total internal reflection occurs and no light escapes to illuminate the ceiling outside the ring.

The critical angle is given by:

$\sin \theta_c = \dfrac{1}{\mu} = \dfrac{1}{4/3} = \dfrac{3}{4}$

From the geometry of the container, the maximum radius $r$ of the ring is related to the depth of the water $h$ by:

$\tan \theta_c = \dfrac{r}{h}$

Using $\sin \theta_c = \dfrac{3}{4}$, we find $\tan \theta_c$:

$\tan \theta_c = \dfrac{3}{\sqrt{4^2 - 3^2}} = \dfrac{3}{\sqrt{7}}$

Given $h = 20 \text{ cm}$, the maximum radius is:

$r = h \tan \theta_c = 20 \times \dfrac{3}{\sqrt{7}} = \dfrac{60}{\sqrt{7}} \text{ cm}$

Calculating the numerical value:

$r \approx \dfrac{60}{2.646} \approx 22.68 \text{ cm}$

Rounding to one decimal place, we get $22.7 \text{ cm}$.