$A$ planoconvex lens,when silvered at its plane surface,is equivalent to a concave mirror of focal length $28 \ cm$. When its curved surface is silvered and the plane surface is not silvered,it is equivalent to a concave mirror of focal length $10 \ cm$. The refractive index of the material of the lens is:

$A$ plano-convex lens $(f = 20 \ cm)$ is silvered at the plane surface. Now,the new focal length $F$ will be........$cm$.

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

An upright object is placed at a distance of $40\, cm$ in front of a convergent lens of focal length $20\, cm$. $A$ convergent mirror of focal length $10\, cm$ is placed at a distance of $60\, cm$ on the other side of the lens. The position and size of the final image will be

$A$ point source is placed $20 \,cm$ to the left of a concave lens of focal length $10 \,cm$.
$(a)$ Where is the image formed?
$(b)$ Where to the right of the lens would you place a concave mirror of focal length $5 \,cm$,so that the final image is coincident with the source?
$(c)$ Where would the final image be formed,if the concave mirror is replaced by a plane mirror at the same position?

$A$ point object is held above a thin equiconvex lens at its focus. The focal length is $0.1 \ m$ and the lens rests on a horizontal thin plane mirror. The final image will be formed at