$A$ narrow parallel beam of light is incident paraxially on a solid transparent sphere of radius $r$ kept in air. What should be the refractive index if the beam is to be focused at the farther surface of the sphere?

$A$ point source of light at the surface of a sphere causes a parallel beam of light to emerge from the opposite surface of the sphere. The refractive index of the material of the sphere is

$A$ 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 $R$)

Refer to the figure given below. $\mu_1$ and $\mu_2$ are the refractive indices of air and the lens material,respectively. The height of the image will be . . . . . . cm.

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea $(7.8 \, mm)$. This surface separates two media of refractive indices $1$ and $1.34$. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus in $cm$.

$A$ clear transparent glass sphere $(\mu=1.5)$ of radius $R$ is immersed in a liquid of refractive index $1.25$. $A$ parallel beam of light incident on it will converge to a point. The distance of this point from the center will be