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?

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,
\[\frac{\sin i}{\sin r} = \frac{1}{\mu_w}\]
\[\frac{\sin i}{\sin r} = \frac{3}{4}\]
\[As r = 45^\circ \]
\[ \Rightarrow \sin i = \frac{3}{4\sqrt{2}}\]
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.

\[\tan  45^\circ= \frac{x + 10}{d}\] 

\[ \Rightarrow d = x + 10           .  .  . (i)\] 

\[and  \] 

\[\tan  i = \frac{x}{d}\] 

\[ \Rightarrow \frac{3}{\sqrt{23}} = \frac{d - 10}{d}  \left( \because \sin  i = \frac{3}{4\sqrt{2}} \Rightarrow \tan  i = \frac{3}{\sqrt{23}} \right)\] 

\[ \Rightarrow \frac{3}{\sqrt{23}} - 1 = \frac{- 10}{d}\] 

\[ \Rightarrow d = \frac{\sqrt{23} \times 10}{\sqrt{23} - 3}\] 

\[d = 26 . 7  cm\]
Hence, the required minimum height of water = 26.7 cm