Question

A cylindrical vessel of diameter 12 cm contains 800π cm³ 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 vessel under the glass piece is seen by the paraxial rays (see figure), locate its image. The index of refraction of glass is 1.50 and that of water is 1.33.
Figure

Answer 1

Given,
Diameter of the cylindrical vessel = 12 cm
∴ radius r = 6 cm
Diameter of the cylindrical glass piece = 8 cm
∴ radius r_^' = 4 cm and its height, h₁ = 8 cm
Refractive index of glass, μ_g = 1.5
Refractive index of water, μ_w = 1.3
Let h be the final height of the water column.


The volume of the cylindrical water column after the glass piece is put will be:
$\pi$r²h = 800π + $\pi$r'²h₁
or r²h = 800 + r_'²h₁
or (6)² h = 800 + (4)² 8
$\mathrm{or}h=\frac{800+128}{36}=\frac{928}{36}=25.7\mathrm{cm}$

There will be two shifts; due to the glass block as well as water:

$$\begin{gathered} ∆t_1=\left[1-\frac{1}{{\mu }_g}\right]t_g=\left[1-\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]\times 8=2.26\mathrm{cm} \\ ∆t_2=\left[1-\frac{1}{{\mu }_{\mathrm{w}}}\right]t_{\mathrm{w}}=\left[1-\frac{1}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right]\times \left(25.7-8\right)=4.44\mathrm{cm} \end{gathered}$$

Hence, the total shift = (Δt₁+ Δt₂) = (2.66 + 4.44) cm = 7.1 cm above the bottom.