Question

Water (with refractive index $=4/3)$ in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature $R=6cm$ as shown. Consider oil to act as a thin lens. An object S is placed 24 cm above water surface. The location of its image is at 'x' cm above the bottom of the tank. Then x is

Answer 1

Refraction from a spherical surface is given by,

$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

For refraction from the first surface,

$\frac{7}{4v_1}-\frac{1}{-24}=\frac{\frac{3}{4}}{6}$

For refraction from second surface,

$\frac{4}{3v_2}-\frac{7}{4v_1}=\frac{\frac{4}{3}-\frac{7}{4}}{\infty }$

Adding the two equations gives,

$v_2=16cm$

Thus distance from the bottom=$depth-v_2=18cm-16cm=2cm$