Question

A small object is kept at the centre of bottom of cylindrical beaker of diameter $6cm$ and height $4cm$ filled completely with water $(\mu =\frac{4}{3})$. Consider the light ray from object leaving the beaker through a corner. If this ray and the ray along the axis of beaker is used to locate the image, find the apparent depth in this case.

Answer 1

Due to refraction image is formed at $I$.

$OP=$ depth of water $=4cm$

$PQ=$ radius $=3cm$

$\mu \sin i=1\sin r$

$\therefore \frac{4}{3}=1\sin r$

$\frac{4}{3}\times \frac{3}{5}=\sin r$

$⟹\sin r=\frac{4}{5}⟹r={53}^o$

$\therefore x=3\tan (90-r)$

$=3\tan {37}^o$

$=3\times \frac{3}{4}$

$=\frac{9}{4}$

$\therefore$ apparent depth $=\frac{9}{4}cm$