$A$ 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 $cm$)

The following graph depicts the inverse of magnification $(1/m)$ versus the distance between the object and the lens $(u)$ for a setup. The focal length of the lens used in the setup is .................... $m$.

Inside a vessel filled with liquid, a converging lens is placed as shown in the figure. The lens has a focal length of $15 \,cm$ when in air and has a refractive index of $\frac{3}{2}$. If the liquid has a refractive index of $\frac{9}{5}$, the focal length of the lens in the liquid is (in $\,cm$)

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 $cm$)

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

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?