The position of the final image formed by the given lens combination from the third lens will be at a distance of $(f_1 = +10 \ cm, f_2 = -10 \ cm, f_3 = +30 \ cm)$.

An eye specialist prescribes spectacles having a combination of a convex lens of focal length $40\, cm$ in contact with a concave lens of focal length $25\, cm$. The power of this lens combination in diopters is

$A$ convergent lens is placed $40 \,cm$ to the right of a diverging lens of focal length $15 \,cm$. $A$ parallel beam of light enters the diverging lens from the left, and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is (in $\,cm$)

$A$ biconvex glass lens has a focal length $f$ and power $P$. It is cut into two symmetric parts along its principal axis. If these two parts are joined as shown in figure $(b)$,what is the new power of the combination?

The following figure represents two biconvex lenses $L_1$ and $L_2$ having focal lengths $10 \,cm$ and $15 \,cm$ respectively. The distance between $L_1$ and $L_2$ is: (in $\,cm$)

Two lenses of power $6D$ and $-2D$ are combined to form a single lens. The focal length of this lens will be (in $m$)