$A$ symmetric thin biconvex lens is cut into four equal parts by two planes $AB$ and $CD$ as shown in the figure. If the power of the original lens is $4 \text{D}$, then the power of one part of the divided lens is: (in $\text{D}$)

$A$ thin converging lens of focal length $f=25 \,cm$ forms the image of an object on a screen placed at a distance of $75 \,cm$ from the lens. The screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object has to be shifted, so that its image on the screen is sharp again, is (in $\,cm$)

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $3\;m$ away by means of a large convex lens. What is the maximum possible focal length (in $m$) of the lens required for the purpose?

$A$ converging lens of focal length $20\, cm$ and diameter $5\, cm$ is cut along the line $AB$. The part of the lens shown shaded in the diagram is now used to form an image of a point $P$ placed $30\, cm$ away from it on the line $XY$,which is perpendicular to the plane of the lens. The image of $P$ will be formed:

$A$ 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 $cm$)

$A$ double convex lens of focal length $20 \ cm$ is made of glass of refractive index $3/2$. When placed completely in water $(_{a}\mu_{w} = 4/3)$,its focal length will be.....$cm$.