$A$ screen is placed $90 \ cm$ from an object. The image is formed on the screen by a convex lens placed at two different locations separated by $20 \ cm$. The focal length of the lens is approximately equal to: (in $cm$)

The image distance of an object placed $10 \ cm$ in front of a thin lens of focal length $+ 5 \ cm$ is......$cm$.

The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index $= 1.7$) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of the radius of curvature of the front and back surface of the biconvex lens is . . . . . . .

If ${I_1}$ and ${I_2}$ are the sizes of the images for the two positions of the lens in the displacement method,then the size of the object is given by:

Two thin lenses having $R_1$ and $R_2$ as the radii of curvature of their surfaces are kept coaxially together. Their power is proportional to

$A$ beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \, cm$ from $P$. At what point does the beam converge if the lens is a concave lens of focal length $16 \, cm$ (in $, cm$)?