At what distance from a biconvex lens of focal length $F$ must an object be placed so that the distance between the object and its real image is minimal?

When an object of height $12 \ cm$ is placed at a distance from a convex lens, an image of height $18 \ cm$ is formed on a screen. Without changing the positions of the object and the screen, if the lens is moved towards the screen, another clear image is formed on the screen. The height of this image is (in $cm$)

$A$ 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).

The size of the images of an object, formed by a thin lens, are equal when the object is placed at two different positions $8 \ cm$ and $24 \ cm$ from the lens. The focal length of the lens is . . . . . . $cm$.

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. If the central region of the lens with aperture diameter $\frac{d}{2}$ is covered by black paper,what will be the new focal length and intensity of the image?

There is an equiconvex glass lens with radius of each face as $R$,$_a{\mu _g} = 3/2$,and $_a{\mu _w} = 4/3$. If there is water in the object space and air in the image space,then the focal length is: