Question

A transparent solid sphere of radius $2\text{cm}$ and density $\rho$ floats in a transparent liquid of density $2\rho$ kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature $8\text{cm}$ and is silvered to make it a concave mirror as shown in the figure. When an object is placed at a distance of $10\text{cm}$ directly above the centre of the sphere $C$, its final image coincides with it. Find approximate value of $h$, the height of the liquid surface in the beaker from the apex of the bottom. Consider the paraxial rays only. The refractive index of the sphere is $\frac{3}{2}$ and that of the liquid is $\frac{4}{3}$.

• A. $17\text{cm}$
• B. $11\text{cm}$
• C. $13\text{cm}$
• D. $15\text{cm}$

Answer 1

The correct option is D $15\text{cm}$

Let, volume of the solid $=V_s$;
volume of the liquid displaced$=V_L$

From equilibrium of forces on the sphere,
$F_B={\rho }_s{V}_{s}g$

$\Rightarrow {\rho }_L{V}_{L}g={\rho }_s{V}_{s}g$

$\Rightarrow \frac{V_s}{V_L}=\frac{{\rho }_L}{{\rho }_s}$

$\Rightarrow \frac{V_s}{V_L}=\frac{2\rho }{\rho }=2$

$\Rightarrow V_s=2V_L$

This means half the sphere is inside the liquid.

For the image to coincide with the object, light should fall normally on the bottom spherical surface. For this, the image formed by transparent solid sphere should be at the centre of curvature of the bottom spherical surface.

From lens makers formula at sphere-air interface,

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}.......\left(1\right)$

where,
$u=-(10-2)\text{cm}=-8\text{cm}$

${\mu }_2=$refractive index of solid sphere $=\frac{3}{2}$

${\mu }_1=$refractive index of air $=1$,

Substituting the values in $\left(1\right)$,

$\Rightarrow \frac{3/2}{v_1}-\frac{1}{(-8)}=\frac{3/2-1}{+2}$

$\Rightarrow v_1=12\text{cm}$


Further applying the lens makers' formula for sphere-liquid interface,

$\Rightarrow \frac{4/3}{v}-\frac{3/2}{8}=\frac{4/3-3/2}{-2}$

$\Rightarrow v=\frac{64}{13}=4.92\text{cm}$

So,
$h=CB+B{C}^{\prime }+C^{\prime }P$

$\Rightarrow h=2+4.92+8$

$\therefore h\approx 15\text{cm}$