The power of an equi-concave lens is $-4.5 \ D$ and it is made of a material with a refractive index of $1.6$. The radius of curvature of the lens is: (in $cm$)

$A$ convex lens of focal length $\frac{1}{3} \ m$ forms a real,inverted image twice the size of the object. The distance of the object from the lens is (in $m$)

If sunlight is focused on a paper using a convex lens,it starts burning the paper in the shortest time when the lens is kept at $30 \text{ cm}$ above the paper. If the radius of curvature of the lens is $60 \text{ cm}$,then the refractive index of the lens material is $\frac{\alpha}{10}$. The value of $\alpha$ is . . . . . . .

$A$ 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:

$A$ lens with a focal length of $50 \ cm$ forms an image of a distant object that subtends an angle of $1 \text{ milliradian}$ at the lens. What is the size of the image in $mm$?

$A$ magnifying glass is to be used at a fixed object distance of $1 \text{ inch}$. If it is to produce an erect image magnified $5$ times,its focal length should be ........ $\text{inch}$.