$A$ beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \, cm$ from $P$. At what point does the beam converge if the lens is a concave lens of focal length $16 \, cm$ (in $, cm$)?

The distance between an object and the screen is $100 \ cm$. $A$ lens produces an image on the screen when placed at either of the positions $40 \ cm$ apart. The power of the lens is

The power of an equi-concave lens is $-4.5 \ D$ and it is made of a material with a refractive index of $1.6$. The radius of curvature of the lens is: (in $cm$)

An object is located in a fixed position in front of a screen. $A$ sharp image is obtained on the screen for two positions of a thin lens separated by $10\, cm$. The size of the images in the two situations are in the ratio $3 : 2$. What is the distance between the screen and the object in $cm$?

The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to (in $\,cm$)

Two point sources $S_1$ and $S_2$ 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 (in $cm$)?