$A$ convex lens:

An object is placed at a distance $m$ times the focal length of a divergent lens. The size of the image is shorter than that of the object by what factor?

$A$ glass convex lens $(\mu_{g}=1.5)$ has a focal length of $8 \ cm$ when placed in air. What would be the focal length of the lens in $cm$ when it is immersed in water $(\mu_{w}=\frac{4}{3})$?

The power (in diopters) of an equiconvex lens with radii of curvature of $10 \ cm$ and refractive index of $1.6$ is

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 $cm$)?

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?