An object is located $4 \mathrm{m}$ from the first of two thin converging lenses of focal lengths $2 \mathrm{m}$ and $1 \mathrm{m}$ respectively. The lenses are separated by $3 \mathrm{m}$. The final image formed by the second lens is located from the source at a distance of

A plano-convex lens fits exactly into plano-concave lens as shown in the figure. Their plane surfaces are parallel to each other. If the lenses are made of different materials of refractive indices $1.6$ and $1.5$ respectively, if $R$ is the radius of curvature of the curved surface of lenses, then the focal length of the combination

Two identical equiconvex lenses, each of focal length $f$ are placed side by side in contact with each other with a layer of water in between them as shown in the figure. If refractive index of the material of the lenses is greater than that of water, how the combined focal length $F$ is related to $f$?

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surfaces are of the same radius of curvature $R=14 \mathrm{cm}.$ For this bi-convex lens, for an object distance of $40 \mathrm{cm},$ the image distance will be