$A$ 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 $R$)

In a medium of refractive index $1.6$ having a convex surface,a point object is placed at a distance of $12 \,cm$ from the pole. The radius of curvature of the surface is $6 \,cm$. Locate the image as seen from air.

$A$ clear transparent glass sphere $(\mu=1.5)$ of radius $R$ is immersed in a liquid of refractive index $1.25$. $A$ parallel beam of light incident on it will converge to a point. The distance of this point from the center will be

$A$ parallel beam of light is allowed to fall on a transparent spherical globe of diameter $30 \,cm$ and refractive index $1.5$. The distance from the centre of the globe at which the beam of light can converge is . . . . . . $mm$.

Write the equation for the image formed by a curved surface of radius of curvature $R$ in a medium of refractive index $n_1$.

$A$ spherical surface of radius of curvature $R$ separates air from glass (refractive index $= 1.5$). The centre of curvature is in the glass medium. $A$ point object $O$ is placed in air on the optic axis of the surface,so that its real image is formed at $I$ inside the glass. The line $OI$ intersects the spherical surface at $P$ and $PO = PI$. The distance $PO$ is equal to: (in $R$)