Double convex lenses are to be manufactured from a glass of refractive index $1.55$ with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \ cm$ (in $cm$)?

Focal length of a convex lens is $20 \ cm$ and its refractive index is $1.5$. It produces an erect,enlarged image if the distance of the object from the lens is: (in $cm$)

$A$ 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 $\,cm$)

Two transparent media of refractive indices $\mu_1$ and $\mu_3$ have a solid lens-shaped transparent material of refractive index $\mu_2$ between them as shown in the figures in Column $II$. $A$ ray traversing these media is also shown in the figures. In Column $I$,different relationships between $\mu_1, \mu_2$,and $\mu_3$ are given. Match them to the ray diagram shown in Column $II$.

If the radius of curvature of the convex surface of a plano-convex lens is $10 \ cm$ and its focal length is $30 \ cm$,what is the refractive index of the material of the lens?

The two surfaces of a concave lens, made of glass of refractive index $1.5$, have the same radii of curvature $R$. It is now immersed in a medium of refractive index $1.75$. Then the lens: