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 $cm$.

$A$ ray coming from an object which is situated at infinity in the air falls on a spherical glass surface $(n=1.5)$. The distance of the image will be . . . . . . . ($R$ is the radius of curvature of the spherical glass surface.)

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

$A$ glass rod has ends as shown in the figure. The refractive index of glass is $\mu$. The object $O$ is at a distance $2R$ from the surface of larger radius of curvature. The distance between the apexes of the ends is $3R$. The range of $\mu$ for which the image is real is given by

$A$ fish is at the centre of a spherical water-filled $(\mu = 4/3)$ fish bowl. $A$ child stands in air at a distance $2R$ ($R$ is the radius of curvature of the sphere) from the centre of the bowl. At what distance from the centre would the child's nose appear to the fish situated at the centre (in $R$)?

The slab of a material of refractive index $2$ shown in the figure has a curved surface $APB$ of radius of curvature $10 \, cm$ and a plane surface $CD$. On the left of $APB$ is air and on the right of $CD$ is water with refractive indices as given in the figure. An object $O$ is placed at a distance of $15 \, cm$ from the pole $P$ as shown. The distance of the final image of $O$ from $P$,as viewed from the left,is ..... $cm$.