The power (in diopters) of an equiconvex lens with radii of curvature of $10 \ cm$ and refractive index of $1.6$ is

$A$ double convex lens,made of a material of refractive index $\mu_1$,is placed inside two liquids of refractive indices $\mu_2$ and $\mu_3$ as shown,where $\mu_2 > \mu_1 > \mu_3$. $A$ wide,parallel beam of light is incident on the lens from the left. The lens will give rise to:

If an object is placed at $A$ $(OA > f)$,where $f$ is the focal length of the lens,the image is found to be formed at $B$. $A$ perpendicular is erected at $O$ and $C$ is chosen on it such that the angle $\angle BCA$ is a right angle. Then the value of $f$ will be

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is

$A$ plano-convex lens is made of refractive index $1.6$. The radius of curvature of the curved surface is $60 \ cm$. The focal length of the lens is $..... \ cm$.

$A$ thin equiconvex lens is made of glass of refractive index $1.5$ and its focal length is $0.2 \ m$. If it acts as a concave lens of $0.5 \ m$ focal length when dipped in a liquid,the refractive index of the liquid is: