In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?

The distance between an object and the screen is $80 \, cm$. $A$ lens produces an image on the screen when the lens is placed at either of the positions $20 \, cm$ apart. The focal length of the lens is ....... $cm$.

When will the convergent nature of a convex lens be less as compared to its nature in air?

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

The distance between an object and its $3$ times magnified virtual image produced by a convex lens is $20 \,cm$. The focal length of the lens used is: (in $\,cm$)

An object is placed in front of a convex lens,and an image of height $8 \, cm$ is formed. When the lens is moved to a new position,the image height becomes $2 \, cm$. Find the height of the object in $cm$.