The formation of a real image using a biconvex lens is shown below:
If the whole setup is immersed in water without disturbing the object and the screen positions,what will one observe on the screen?

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 $cm$.

$A$ glass convex lens has a refractive index of $1.55$ with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \,cm$ (in $\,cm$)?

$A$ turnip sits before a thin converging lens,outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center [as indicated in figure $(a)$],you can change the curvature of its front and rear sides. What happens to the lateral height of the image?

The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 \ cm$. The lens is now immersed in a liquid of refractive index $1.5$. The ratio of power of the lens in air to its power in the liquid will be $x : 1$. The value of $x$ is $.....$