A cylindrical vessel of diameter 12 cm contains$800\pi {\text{cm}}^3$ of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0 cm is placed in the vessel. If the bottom of the vesel under the glass piece is seen by the paraxial rays, locate its image. The index of refraction of glass is 1.50 and that of water is1.33.

Given $\text{r}=6{\text{cm,r}}_{\text{1}}=4{\text{cm,h}}_{\text{1}}=8\text{cm}$

Let, h = Final height of water column

The volume of the cylindrical water column after the glass piece is put will be,

$\pi r^2\text{h}=800\pi +\pi {\text{r}}_{\text{1}}^{\text{2}}{\text{h}}_{\text{1}}$

$\left(or\right){\text{r}}^{\text{2}}\text{h}=800+{\text{r}}_{\text{1}}^{\text{2}}{\text{h}}_{\text{1}}$

$\left(or\right)6^2\text{h}=800+4^2\times 8=25.7\text{cm}$                                

There are two shifts due to glass block as well as water

$\text{So,}\Delta {\text{t}}_1=\left(1-\frac{1}{{\mu }_0}\right){\text{t}}_{\text{0}}=\left(1-\frac{1}{\text{3/2}}\right)8=2.2\text{6cm}$

$\text{and,}\Delta {\text{t}}_2=\left(1-\frac{1}{{\mu }_{\text{w}}}\right){\text{t}}_{\text{w}}=\left(1-\frac{1}{\text{4/3}}\right)\left(25.7-8\right)=4.44\text{cm}$

Total shift $\text{=}\left(2.66+4.44\right)\text{cm}=\text{7}\text{.1cm}$ above the bottom