$A$ 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 . . . . . . $mm$.

In a long glass tube,a mixture of two liquids $A$ and $B$ with refractive indices $1.3$ and $1.4$ respectively,forms a convex refractive meniscus towards $A$. If an object placed at $13 \ \text{cm}$ from the vertex of the meniscus in $A$ forms an image with a magnification of $-2$,then the radius of curvature of the meniscus is:

The figure shows a transparent sphere of radius $R$ and refractive index $\mu$. An object $O$ is placed at a distance $x$ from the pole of the first surface so that a real image is formed at the pole of the exactly opposite surface. If $x = 2R$,then the value of $\mu$ is

$A$ 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 $R$)

$A$ concave spherical surface of radius of curvature $10 \ cm$ separates two media $X$ and $Y$ of refractive index $4/3$ and $3/2$ respectively. If the object is placed along the principal axis in medium $X$,then:

$A$ small object is enclosed in a transparent solid sphere of radius $8 \ cm$. The object is situated at $2 \ cm$ from the centre of the sphere. If its image appears to be at $3.2 \ cm$ from the nearest side,then the refractive index of the material of the sphere is: