$A$ 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 refractive index $1.5$ and focal length $18 \, cm$ in air is immersed in water. The change in focal length of the lens will be $........... \, cm$. (Given refractive index of water $= 4/3$)

Look at the ray diagram shown. What will be the focal length of the $1^{st}$ and the $2^{nd}$ lens,if the incident light ray passes without any net deviation?

$A$ point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 \; cm$ and the refractive index of the lens material is $1.5$. Then the focal length of the lens (in $cm$) is:

$A$ thin convex lens made from crown glass $\left( \mu = \frac{3}{2} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$,it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is:

Is power negative or positive for a concave lens?