$A$ 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:

$A$ spherical surface of radius of curvature $R$ separates air (refractive index $1.0$) from glass (refractive index $1.5$). The centre of curvature is in the glass. $A$ point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $PQ$ cuts the surface at a point $O$,and $PO = OQ$. The distance $PO$ is equal to (in $R$)

$A$ luminous point object $O$ is placed at a distance $2R$ from the spherical boundary separating two transparent media of refractive indices $n_1$ and $n_2$ as shown,where $R$ is the radius of curvature of the spherical surface. If $n_1 = \frac{4}{3}$,$n_2 = \frac{3}{2}$ and $R = 10 \text{ cm}$,the image is obtained at a distance from $P$ equal to:

$A$ parallel beam of light travelling in air (refractive index $1.0$) is incident on a convex spherical glass surface of radius of curvature $50 \ cm$. The refractive index of glass is $1.5$. The rays converge to a point at a distance $x \ cm$ from the centre of curvature of the spherical surface. The value of $x$ is . . . . . . $cm$.

$A$ point object is placed at the centre of a glass sphere of radius $6 \, cm$ and refractive index $1.5$. The distance of the virtual image from the surface of the sphere is.....$cm$