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

$A$ parallel beam of light travelling in air (refractive index $1.0$) is incident on a convex spherical glass surface of radius of curvature $50 \ cm$. The refractive index of glass is $1.5$. The rays converge to a point at a distance $x \ cm$ from the centre of curvature of the spherical surface. The value of $x$ is . . . . . . $cm$.

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

$A$ concave spherical refracting surface separates two media,glass and air $(\mu_1 = 1.5, \mu_2 = 1.0)$. If the image is to be real,at what minimum distance $u$ should the object be placed in the glass if $R$ is the radius of curvature?

$A$ hemispherical glass lens with a refractive index of $1.5$ is placed in a liquid with a refractive index of $1.3$ (see the figure). The radius of the hemispherical lens is $10 \,cm$. $A$ parallel beam of light traveling in the liquid is refracted by the glass lens. The absolute value of the position of the image from the center of the glass lens will be (in $\,cm$)

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