$A$ convex lens is immersed in water. How will its power $P$ change?

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

$A$ student measures the focal length of a convex lens by obtaining an image of an object pin at a distance '$v$' from the lens,with the object placed at a distance '$u$' from the lens. What will the graph between '$u$' and '$v$' look like?

The size of the real image produced by a convex lens of focal length $F$ is '$m$' times the size of the object. The image distance from the lens is

$A$ thin convex lens is made of a material of refractive index $1.6$. An object is kept at a distance of $u$ from the lens on the principal axis as shown in the figure. The radii of curvature of the surfaces are $10 \, cm$ and $5 \, cm$. Now,the lens is reversed such that the face having radius of curvature $5 \, cm$ lies close to the object. The difference in image position as obtained for both the cases is equal to ......$u$.

An object is placed at a distance of $f/2$ from a convex lens. The image will be