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

Light from a point source in air falls on a spherical glass surface (refractive index,$\mu=1.5$ and radius of curvature $=50\ cm$). The image is formed at a distance of $200\ cm$ from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is . . . . . . $m$.

$A$ beam of diameter $d$ is incident on a glass hemisphere as shown. If the radius of curvature of the hemisphere is very large in comparison to $d$,then the diameter of the beam at the base of the hemisphere will be:

Locate the image formed by refraction in the situation shown in the figure.

The figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If an object is placed at a distance $R$ from the pole of the first surface,then the real image is formed at a distance $R$ from the pole of the second surface. The refractive index $\mu$ of the sphere is given by

$A$ spherical surface of radius of curvature $R$ separates air (refractive index $1.0$) from glass (refractive index $1.5$). The centre of curvature is in the glass. $A$ point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$,and $PO = OQ$. The distance $PO$ is equal to (in $R$)