When will the convergent nature of a convex lens be less as compared to its nature in air?

The graph between $\frac{1}{u}$ and $\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of the lens is $1.5$ and both its surfaces have the same radius of curvature $R$. The value of $R$ will be in $cm$.
(Where $u =$ object distance,$v =$ image distance)

$A$ biconvex lens of focal length $f$ and radii of curvature of both surfaces $R$ is made of a material of refractive index $n_{1}$. This lens is placed in a liquid of refractive index $n_{2}$. How will this lens behave?

If the image formed by a thin convex lens of power $P$ has magnification $m$,then the image distance $v$ is:

$A$ lens placed $10 \,cm$ away from a wall casts a sharp inverted image of a candle on it. It again casts a sharp image when the lens is moved $20 \,cm$ further away from the wall. Now,the candle and the lens are moved such that a sharp inverted image with unit magnification is formed on the wall. To achieve this configuration,the candle was moved

If in a plano-convex lens,the radius of curvature of the convex surface is $10 \,cm$ and the focal length of the lens is $30 \,cm$,the refractive index of the material of the lens will be ..........