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 $cm$)

$A$ 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 $\,cm$)

Locate the image formed by refraction in the situation shown in the figure.

$A$ 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 $R$)

$A$ 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:

$A$ 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.)