The figure shows an equiconvex lens of focal length $f$. If the lens is cut along $AB$,the focal length of each half will be

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:

$A$ plano-convex lens,when silvered on the plane side,behaves like a concave mirror of focal length $30 \ cm$. However,when silvered on the convex side,it behaves like a concave mirror of focal length $10 \ cm$. The refractive index of its material is:

The diagram shows a silvered equiconvex lens. An object of length $1\,cm$ is placed at a distance of $30\,cm$ in front of the lens. What will be the final image properties? The refractive index of the lens is $\mu$ and the refractive index of the medium in which the lens has been placed is $2\mu$. Both surfaces have a radius of curvature $R$.

An equiconvex lens has power $P$. It is cut into two symmetrical halves by a plane containing the principal axis. The power of one part will be

The distance between a convex lens and a plane mirror is $10 \, cm$. The parallel rays incident on the convex lens after reflection from the mirror form an image at the optical centre of the lens. The focal length of the lens will be.....$cm$.