On which does the convergence capacity of a lens depend?

Two point sources $P$ and $Q$ are $24 \ cm$ apart. Where should a convex lens of focal length $9 \ cm$ be placed in between them so that the images of both sources are formed at the same place?

An object placed at a distance of $16 \ cm$ from a convex lens produces an image of magnification $m$ $(m > 1)$. If the object is moved towards the lens by $8 \ cm$,then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is (in $cm$)

$A$ thin biconvex lens is made of glass $(\mu = 1.50)$ and both surfaces have a radius of curvature of $20 \ cm$. $A$ beam of incident light is parallel to the principal axis of the lens. The lens converges it at a distance $L \ cm$. Then $L = \dots$

An equiconvex lens of radius of curvature $14 \ cm$ is made up of two different materials. The left half and right half of the vertical portion are made up of materials with refractive indices $1.5$ and $1.2$ respectively,as shown in the figure. If a point object is placed at a distance of $40 \ cm$,calculate the image distance. (in $cm$)

$A$ convex lens is made of $3$ layers of glass of $3$ different materials as shown in the figure. $A$ point object is placed on its axis. The number of images of the object formed is: