An object is placed at a distance of $1.50 \, m$ from a screen. $A$ convex lens is placed between them to produce a four times magnified image on the screen. The focal length of the lens is ..... $cm$.

$A$ 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 $cm$)?

$A$ convex lens of radii of curvature $6 \ cm$ and $12 \ cm$ is immersed in a liquid of refractive index $1.3$. If the refractive index of the material of the lens is $1.5$, then the focal length of the lens when immersed in the liquid is (in $cm$)

In the diagram shown,the lens is moving towards the object with a velocity $V \, m/s$ and the object is also moving towards the lens with the same speed. What is the speed of the image with respect to the earth when the object is at a distance $2f$ from the lens? ($f$ is the focal length.)

Double convex lenses are to be manufactured from a glass of refractive index $1.55$ with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \ cm$ (in $cm$)?

The focal length of a lens with a refractive index of $1.5$ in air is $0.30 \ m$. When it is immersed in a liquid of refractive index $\frac{4}{3}$,its focal length in the liquid will be ...... $cm$.