$A$ convex lens of focal length $f$ $m$ forms a real,inverted image twice the size of the object. The object distance from the lens in metres 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 $\,cm$)

$A$ convex lens of focal length $20 \ cm$ produces images of the same magnification $2$ when an object is kept at two distances $x_1$ and $x_2$ $(x_1 > x_2)$ from the lens. The ratio of $x_1$ and $x_2$ is

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?

An object placed at a distance of $16 \ cm$ from a convex lens produces an image of magnification $m$ $(m > 1)$. If the object is moved towards the lens by $8 \ cm$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is (in $cm$)

$A$ turnip sits before a thin converging lens,outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center [as indicated in figure $(a)$],you can change the curvature of its front and rear sides. When you squeeze the lens,the image: