$A$ bulb is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance $d$ ahead of the second wall. The required focal length will be:

$A$ convex lens is made from glass material having a refractive index of $1.4$ with the same radius of curvature on both sides. The ratio of its focal length to its radius of curvature is . . . . . . .

An extended object is placed at point $O$,$10 \ cm$ in front of a convex lens $L_1$ and a concave lens $L_2$ is placed $10 \ cm$ behind it,as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are $20 \ cm$. The refractive index of both the lenses is $1.5$. The total magnification of this lens system is

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 $cm$)

$A$ lens of refractive index '$\mu$' has focal length '$f$'. When the lens is immersed in a liquid of refractive index '$\mu_0$',its focal length becomes '$f_0$'. Then '$f_0$' is given by

$A$ plano-convex lens has a diameter of $6\, cm$ and a thickness of $3\, mm$ at the center. If the speed of light in the material of the lens is $2 \times 10^{8}\, m/s$,then the focal length of the lens is ..... $cm$.