In the displacement method,a convex lens forms a real image of an object for its two different positions. If the heights of the images in the two cases are $24 \,cm$ and $6 \,cm$,then the height of the object is ........ $cm$.

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \ m/s$ and stops at the focus. The image :-

The focal length of a convex lens of refractive index $1.5$ is $2 \ cm$. What will be the focal length of the lens when it is immersed in a liquid of refractive index $1.25$?

$A$ 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 refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

The focal length of a converging lens in air is $R$. If it is dipped in water of refractive index $1.33$,then its focal length will be around (Refractive index of lens material is $1.5$).