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 $cm$)

Refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \text{ cm}$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is

For an object placed at a distance $2.4\,m$ from a lens,a sharp focused image is observed on a screen placed at a distance $12\,cm$ from the lens. $A$ glass plate of refractive index $1.5$ and thickness $1\,cm$ is introduced between the lens and the screen such that the glass plate faces are parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen (in $,m$)?

The slit of a collimator is illuminated by a source as shown in the adjoining figures. The distance between the slit $S$ and the collimating lens $L$ is equal to the focal length of the lens. The correct direction of the emergent beam will be as shown in figure:

Focal length of a convex lens will be maximum for

$A$ glass convex lens has a refractive index of $1.55$ with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \,cm$ (in $\,cm$)?