$A$ convex lens has power $P$. It is cut into two halves along its principal axis. Further,one piece (out of two halves) is cut into two halves perpendicular to the principal axis as shown in the figure. Choose the incorrect option for the reported lens pieces.

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:

The radius of curvature of the convex surface of a plano-convex lens is $12 \ cm$ and its refractive index is $1.5$. Find the focal length of this lens in $cm$ when the plane surface is silvered.

$A$ concave mirror of focal length $f_{1}$ is placed at a distance $d$ from a convex lens of focal length $f_{2}$. $A$ parallel beam of light coming from infinity parallel to the principal axis falls on the convex lens and then after refraction falls on the concave mirror. If the light is to retrace its path,the distance $d$ should be:

$A$ point object is placed at a distance of $20 \ cm$ from a thin plano-convex lens of focal length $15 \ cm$. If the plane surface is silvered,then the image formed is:

$A$ 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?