$A$ point object is moving at a constant speed of $1 \,ms^{-1}$ along the principal axis of a convex lens of focal length $10 \,cm$. The speed of the image is also $1 \,ms^{-1}$, when the object is at what distance (in $cm$) from the optical centre of the lens?

$A$ 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 $cm$)?

What is the relative decrease in focal length of a lens for an increase in optical power by $0.1 \ D$ from $2.5 \ D$? [$D$ stands for dioptre]

The dispersive powers of the glasses of lenses used in an achromatic pair are in the ratio $5 : 3$. If the focal length of the concave lens is $15 \ cm$,then the nature and focal length of the other lens would be:

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?

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 $cm$?