$A$ double convex lens of glass of $\mu = 1.5$ has a radius of curvature of each of its surfaces equal to $0.2 \ m$. The power of the lens is:

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

The figure given below 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 $O$ 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.....$cm$.

At what distance from a biconvex lens of focal length $F$ must an object be placed so that the distance between the object and its real image is minimal?

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

The diameter of the aperture of a plano-convex lens is $6 \, cm$ and its maximum thickness is $3 \, mm$. If the velocity of light in the material of the lens is $2 \times 10^8 \, m/s$,its focal length is ........ $cm$.