The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 \ cm$. The lens is now immersed in a liquid of refractive index $1.5$. The ratio of power of the lens in air to its power in the liquid will be $x : 1$. The value of $x$ is $.....$

Parallel rays are incident on a thick plano-convex lens having radius of curvature $R$,refractive index $\mu$ and thickness $t$. When rays are incident on the plane surface,they converge at a distance $x$ from the plane surface. When rays are incident on the curved surface,they converge at a distance $y$ from the curved surface. Then:

$A$ thin glass (refractive index $1.5$) lens has an optical power of $-5 D$ in air. Its optical power in a liquid medium with a refractive index of $1.6$ will be:

$A$ convex lens of radii of curvature $6 \ cm$ and $12 \ cm$ is immersed in a liquid of refractive index $1.3$. If the refractive index of the material of the lens is $1.5$, then the focal length of the lens when immersed in the liquid is (in $cm$)

$A$ hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids $L_1$ and $L_2$ having refractive indices $n_1$ and $n_2$ respectively $(n_2 > n_1 > 1)$. The lens will diverge a parallel beam of light if it is filled with

Two point sources $P$ and $Q$ are $24 \ cm$ apart. Where should a convex lens of focal length $9 \ cm$ be placed in between them so that the images of both sources are formed at the same place?