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 ' $O$ ' placed in air on the optic axis of the surface, so that its real image is formed at 'I' inside glass. The line OI intersects the spherical surface at $P$ and $P O=P I$. The distance $P O$ equals to

Given a thin convex lens (refractive index $\mu_2$ ), kept in a liquid (refractive index $\mu_1, \mu_1<\mu_2$ ) having radii of curvatures $\left|R_1\right|$ and $\left|R_2\right|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?

A symmetric thin biconvex lens is cut into four equal parts by two planes $A B$ and $C D$ as shown in figure. If the power of original lens is 4D then the power of a part of the divided lens is

In the diagram given below, there are three lenses formed. Considering negligible thickness of each of them as compared to $\left|R_1\right|$ and $\left|R_2\right|$, i.e., the radii of curvature for upper and lower surfaces of the glass lens, the power of the combination is