Look at the ray diagram shown. What will be the focal length of the $1^{st}$ and the $2^{nd}$ lens,if the incident light ray passes without any net deviation?

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is

If a concave lens is placed in the path of converging rays,a real image will be produced if the distance of the pole from the point of convergence of incident rays lies between ($f =$ magnitude of focal length of lens).

Sunlight of intensity $1.3 \text{ kW m}^{-2}$ is incident normally on a thin convex lens of focal length $20 \text{ cm}$. Ignore the energy loss of light due to the lens and assume that the lens aperture size is much smaller than its focal length. The average intensity of light,in $\text{kW m}^{-2}$,at a distance $22 \text{ cm}$ from the lens on the other side is. . . . .

The following figure shows a beam of light converging at point $P$. When a concave lens of focal length $16 \,cm$ is introduced in the path of the beam at a place shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to (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