An equiconvex lens of glass of focal length $0.1 \text{ m}$ is cut along a plane perpendicular to the principal axis into two equal parts. The ratio of the focal lengths of the new lenses formed is

The two surfaces of a biconvex lens have the same radii of curvature. This lens is made of glass of refractive index $1.5$ and has a focal length $10 \ cm$ in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index $= 4/3$), its focal length (in $cm$) is:

An equiconvex lens of radius of curvature $R = 20 \, cm$ and refractive index $\mu = 1.5$ is silvered on one side. The equivalent focal length of the system is: (in $, cm$)

$A$ convex lens of focal length $15\,cm$ is placed in front of a convex mirror. When an object is placed at $O$,the image coincides with it. The radius of curvature of the mirror is.....$cm$.

The plane surface of a plano-convex lens of focal length $f$ is silvered. It will behave as

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