$A$ thin convex lens of refractive index $1.5$ has a focal length of $15\, cm$ in air. When the lens is placed in a liquid of refractive index $4/3$,its focal length will be......$cm$.

An object and a screen are fixed on the uprights of an optical bench. The distance between them is $100 \, cm$. $A$ convex lens is placed in between the object and the screen and the position of the lens is so adjusted that the image of the object is formed on the screen at two conjugate positions of the lens. The distance between these conjugate positions of the lens is $40 \, cm$. What is the focal length of the lens in $cm$?

The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$. What is the focal length of the lens used?

The focal length of a lens with a refractive index of $1.5$ in air is $0.30 \ m$. When it is immersed in a liquid of refractive index $\frac{4}{3}$,its focal length in the liquid will be ...... $cm$.

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it, then the focal length of the plano-convex lens is ($\mu =$ the refractive index of the material of the lens).

$f_V$ and $f_R$ are the focal lengths of a convex lens for violet and red light respectively,and $F_V$ and $F_R$ are the focal lengths of a concave lens for violet and red light respectively. Then: