$A$ concave spherical refracting surface separates two media,glass and air $(\mu_1 = 1.5, \mu_2 = 1.0)$. If the image is to be real,at what minimum distance $u$ should the object be placed in the glass if $R$ is the radius of curvature?

$A$ glass rod has ends as shown in the figure. The refractive index of glass is $\mu$. The object $O$ is at a distance $2R$ from the surface of larger radius of curvature. The distance between the apexes of the ends is $3R$. The range of $\mu$ for which the image is real is given by

$A$ point object $O$ is placed on the axis of a cylindrical piece of glass of refractive index $1.6$ as shown in the figure. One surface of the glass piece is convex with a radius of curvature $3 \,mm$. The point appears to be at $5 \,mm$ on the axis when viewed along the axis from the right side of the convex surface. The distance of the point object from the convex surface is: (in $\,mm$)

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 an object is placed at a distance $R$ from the pole of the first surface,then the real image is formed at a distance $R$ from the pole of the second surface. The refractive index $\mu$ of the sphere is given by

$A$ 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.....$cm$

$A$ spherical interface of radius $R$ separates two media of refractive indices $1$ and $1.4$ respectively, as shown in the figure below. $A$ point source is placed at a distance of $4R$ in front of the spherical interface. The magnitude of the lateral magnification of the point source image is . . . . . . .