A cylindrical vessel of diameter 12cm contains $800\pi\text{cm}^3$ of water. A cylindrical glass piece of diameter 8.0 cm and height 8.0cm is placed in the vessel. If the bottom of the vessel 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 is 1.33.
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Given $r = 6 cm, r_1 = 4cm, h_1 = 8cm$
Let, h = final height of water column.
The volume of the cylindrical water column after the glass piece is put will be,
$\pi\text{r}^2\text{h}=800\pi+\pi\text{r}_1{^2\text{h}_1}$
or $\text{r}^2\text{h}=800+\text{r}_1{^2\text{h}_1}$
or $6^2\text{h}=800+4^2\times8$
$\text{h}=\frac{800+128}{36}=\frac{928}{36}$
$=25.7\text{cm}$
There are two shifts due to glass block as well as water.
So, $\Delta\text{t}_1=\Big(1-\frac{1}{\mu_0}\Big)\text{t}_0=\bigg(1-\frac{1}{\frac{3}{2}}\bigg)8=2.26\text{cm}$
And, $\Delta\text{t}_2=\Big(1-\frac{1}{\mu_{\text{w}}}\Big)\text{t}_{\text{w}}=\bigg(1-\frac{1}{\frac{4}{3}}\bigg)(25.7-8)=4.44\text{cm}$
Total shift = (2.66 + 4.44)cm = 7.1cm above the bottom.
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