$A$ 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:

$A$ narrow parallel beam of light falls on a glass sphere of radius $R$ and refractive index $\mu$ at normal incidence. The distance of the image from the outer edge is given by

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 $cm$.

Derive the relation between object distance $(u)$,image distance $(v)$,refractive indices of the media ($n_1$ and $n_2$),and the radius of curvature $(R)$ for a spherical refracting surface.

An object is placed at a distance of $20 \, cm$ in a rarer medium from the pole of a convex spherical refracting surface of radius of curvature $10 \, cm$. If the refractive index of the rarer medium is $1$ and that of the denser medium is $2$,then the position of the image is at

$A$ point object is kept at $P$ in front of a glass sphere of radius $R$. Its image is formed at $Q$ such that $PO = QO$. The refractive index of the material of the glass sphere is $1.4$. The distance $PO$ is equal to: