A transparent solid sphere of radius 2 cm and density ρ floats in a transparent liquid of density 2 ρ kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature 8 cm and

Question

A transparent solid sphere of radius 2 cm and density ρ floats in a transparent liquid of density 2 ρ kept in a beaker. The bottom of the beaker is spherical in shape with radius of curvature 8 cm and is silvered to make it concave mirror as shown in the figure. When an object is placed at a distance of 10 cm directly above the centre of the sphere its final image coincides with it. Find h (as shown in the figure), 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 3/2 and that of the liquid 4/3 .

TestHub · Accepted Answer

Refraction at surface (1) of the sphere: $\frac{3}{2 v _{1}} + \frac{1}{8} = \frac{1}{4} \Rightarrow v_{1} = 12$ cm.

Refraction at surface (2) (sphere to liquid): $\frac{4}{3 v _{2}} - \frac{3}{2 ( 12 )} = \frac{4/3 - 3/2}{- 2} = \frac{1}{12} \Rightarrow \frac{4}{3 v _{2}} = \frac{1}{16} + \frac{1}{12} \Rightarrow v_{2} = \frac{64}{13}$ cm.

The final image coincides with the object if the concave mirror (R = 8 cm) retraces the path: $h = 2 + v_{2} + 8 = \frac{194}{13}$ cm - (A)

or if the rays converge right at the mirror: $h = 2 + v_{2} = \frac{90}{13}$ cm - (C)

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