Question

A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on a horizontal surface and a small particle P is placed in it at a distance of 5.0 cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible (see figure). The particle P is in the plane of drawing. Up to what minimum height should water be poured in the vessel to make the particle P visible?
Figure

Answer 1

Given,
Diameter and height (h) of the cylindrical vessel = 30 cm
Therefore, its radius (r) = 15 cm
We know the refractive index of water (μ_w) = 1.33 $=\frac{4}{3}$


Using Snell's law,

$$\begin{gathered} \frac{\sin i}{\sin r}=\frac{1}{{\mathrm{\mu }}_{\mathrm{w}}} \\ \frac{\sin i}{\sin r}=\frac{3}{4} \\ \mathrm{As}r=45° \\ \Rightarrow \sin i=\frac{3}{4\sqrt{2}} \end{gathered}$$

Point P will be visible when the refracted ray makes an angle of 45˚ at the point of refraction.
Let x be the distance of point P from X.
$$\begin{gathered} \tan 45°=\frac{x+10}{d} \\ \Rightarrow d=x+10...\left(\mathrm{i}\right) \\ \mathrm{and} \\ \tan \mathrm{i}=\frac{x}{d} \\ \Rightarrow \frac{3}{\sqrt{23}}=\frac{d-10}{d}\left(\because \sin \mathrm{i}=\frac{3}{4\sqrt{2}}\Rightarrow \tan \mathrm{i}=\frac{3}{\sqrt{23}}\right) \\ \Rightarrow \frac{3}{\sqrt{23}}-1=\frac{-10}{d} \\ \Rightarrow d\mathit{=}\frac{\sqrt{23}\times 10}{\sqrt{23}-3} \\ d=26.7\mathrm{cm} \end{gathered}$$

Hence, the required minimum height of water = 26.7 cm