$A$ convex lens has a power of $+5.0 \, D$ in air,where the refractive index of the lens material is $_a\mu_g = 1.5$. In a liquid of what refractive index should it be immersed so that it acts as a concave lens of focal length $100 \, cm$?

$A$ 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 $cm$?

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$,then the ratio of focal lengths of a glass lens in water and in air is

The distance $v$ of the real image formed by a convex lens is measured for various object distances $u$. $A$ graph is plotted between $v$ and $u$. Which one of the following graphs is correct?

What is the relative decrease in focal length of a lens for an increase in optical power by $0.1 \ D$ from $2.5 \ D$? [$D$ stands for dioptre]