The radius of curvature of a convex lens is $40 \,cm$ for each surface. Its refractive index is $1.5$. Its focal length is: (in $\,cm$)

An object $2.4 \ m$ in front of a lens forms a sharp image on a film $12 \ cm$ behind the lens. $A$ glass plate $1 \ cm$ thick,of refractive index $1.50$ is interposed between the lens and the film with its plane faces parallel to the film. At what distance (from the lens) should the object be shifted to be in sharp focus on the film (in $m$)?

What is the focal length (in $cm$) of a biconvex lens with a radius of curvature of $40\, cm$ and a refractive index of $1.65$?

$A$ biconvex lens has a focal length $2/3$ times the radius of curvature of either surface. The refractive index of the material is:

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

$A$ lens is placed between a source of light and a wall. It forms images of area $A_1$ and $A_2$ on the wall for its two different positions. The area of the source of light is