An image is formed at a distance of $100 \ cm$ from the glass surface when light from a point source in air falls on a spherical glass surface with a refractive index of $1.5$. The distance of the light source from the glass surface is $100 \ cm$. The radius of curvature is (in $cm$)

$A$ parallel beam of light emerges from the opposite surface of the sphere when a point source of light lies at the surface of the sphere. The refractive index of the sphere is

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 ..... $cm$.

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 $cm$.

$A$ spherical vessel of radius $10 \, cm$ is filled with water of refractive index $\mu = 4/3$. $A$ fish is at a distance of $4 \, cm$ from the center as shown in the figure. If viewed from end $E$,at what distance will the fish appear to be (in $, cm$)? (Neglect the thickness of the vessel)

$A$ 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: