$A$ point object is placed at the centre of a glass sphere of radius $6 \, cm$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is.....$cm$

Two refracting media are separated by a spherical interface as shown in the figure. $PP'$ is the principal axis, $\mu_1$ and $\mu_2$ are the refractive indices of the medium of incidence and the medium of refraction respectively. Then:

Write the equation for the image formed by a curved surface of radius of curvature $R$ in a medium of refractive index $n_1$.

As shown in the figure,region $BCDEF$ and $ABFG$ have refractive indices $2.0$ and $1.5$ respectively. $A$ particle $O$ is kept at the midpoint of region $BCEF$. The final image of the object as seen by the eye is at a distance of:

In the figure shown here,a point object $O$ is placed in air. $A$ spherical boundary separates various media with a radius of curvature $2.0 \, m$. $AB$ is the principal axis. The refractive index above $AB$ is $1.6$ and below $AB$ is $2.0$. The separation between the images formed due to refraction at the spherical surface is

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