$A$ 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

The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. 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:

$A$ 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 . . . . . . .

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 $,m$)?

$A$ convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black, then

The optical power of a thin glass lens (refractive index $1.5$) in air is $-5 \ D$. What will be its optical power in a liquid with a refractive index of $1.6$ (in $D$)?