$A$ lens behaves as a converging lens in air but a diverging lens in water,then the refractive index $(\mu)$ of its material is

The power of a biconvex lens is $1.25\,m^{-1}$ in a particular medium. The refractive index of the lens is $1.5$ and the radii of curvature are $20\,cm$ and $40\,cm$ respectively. Find the refractive index of the surrounding medium.

An object is placed at a point distant $x$ from the focus of a convex lens and its image is formed at $I$ as shown in the figure. The distances $x$ and $x'$ satisfy the relation:

When an object of height $12 \ cm$ is placed at a distance from a convex lens, an image of height $18 \ cm$ is formed on a screen. Without changing the positions of the object and the screen, if the lens is moved towards the screen, another clear image is formed on the screen. The height of this image is (in $cm$)

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?

$A$ thin convex lens is placed just above an empty vessel of depth $80 \,cm$. The image of a coin kept at the bottom of the vessel is thus formed $20 \,cm$ above the lens. If now water is poured in the vessel up to a height of $64 \,cm$, what will be the approximate new position of the image? Assume that the refractive index of water is $4/3$.