$A$ 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: (in $,D$)

$A$ convex lens of glass $(\mu_g = 1.45)$ has a focal length $f_a$ in air. The lens is immersed in a liquid of refractive index $(\mu_l) = 1.3$. The ratio of the $f_l / f_a$ is

$A$ lens is placed between a source of light and a wall. It forms images of area $A_1$ and $A_2$ on the wall for its two different positions. The area of the source of light is

$A$ plano-concave lens is made of glass of refractive index $1.5$ and the radius of curvature of its curved face is $100 \,cm$. What is the power of the lens?

In the displacement method of a lens,we obtain two images for different positions of the lens. One image is magnified,and the other is diminished. If $m$ is the magnification of the first image,then the focal length of the lens is given by:

$A$ convex lens is made of glass $(\mu_g = 1.5)$ and has a focal length of $4 \, cm$ in air. If it is immersed in water $(\mu_w = 1.33)$,what will be its new focal length in $cm$?