A cylindrical vessel, having both its diameter and height equal to...

Let the bottom-left corner of the vessel be the origin $(0,0)$. The top-right corner is at $(30, 30)$.

The eye is positioned such that the line of sight passes through $(30, 30)$ and $(0,0)$. The angle of refraction $r$ with the vertical is given by $\tan r = \dfrac{30}{30} = 1$, so $r = 45^{\circ}$.

The center of the vessel's bottom is at $x = 15 \text{ cm}$. The particle P is at a distance of $5 \text{ cm}$ from the center. From the figure, it is closer to the origin, so its position is $x = 15 - 5 = 10 \text{ cm}$.

Let water be filled to a height $h$. The ray from P $(10, 0)$ strikes the water surface at $(x_s, h)$ and refracts. To reach the eye, the refracted ray must pass through $(30, 30)$ at an angle of $45^{\circ}$ to the vertical.

The path of the refracted ray in air is a straight line with slope $1$. Thus, $\dfrac{30 - x_s}{30 - h} = \tan 45^{\circ} = 1 \Rightarrow x_s = h$.

In water, the ray travels from $(10, 0)$ to $(h, h)$. The angle of incidence $i$ is given by $\tan i = \dfrac{h - 10}{h}$.

By Snell's law, $\mu \sin i = \sin r$:

$\dfrac{4}{3} \sin i = \sin 45^{\circ} = \dfrac{1}{\sqrt{2}} \Rightarrow \sin i = \dfrac{3}{4\sqrt{2}} = \dfrac{3}{\sqrt{32}}$

From this, $\cos i = \sqrt{1 - \sin^2 i} = \sqrt{1 - \dfrac{9}{32}} = \dfrac{\sqrt{23}}{\sqrt{32}}$.

Therefore, $\tan i = \dfrac{\sin i}{\cos i} = \dfrac{3}{\sqrt{23}}$.

Equating the two expressions for $\tan i$:

$\dfrac{h - 10}{h} = \dfrac{3}{\sqrt{23}}$

$1 - \dfrac{10}{h} = \dfrac{3}{\sqrt{23}}$

$\dfrac{10}{h} = \dfrac{\sqrt{23} - 3}{\sqrt{23}}$

$h = \dfrac{10\sqrt{23}}{\sqrt{23} - 3} = \dfrac{10 \times 4.796}{4.796 - 3} = \dfrac{47.96}{1.796} \approx 26.7 \text{ cm}$.

Answer: $26.7 \text{ cm}$