If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$,then the ratio of focal lengths of a glass lens in water and in air is

$A$ convex lens forming a real image of magnification $m_1$ on a screen is moved through a distance $x$. $A$ new image of magnification $m_2$ is again formed on the screen. The focal length of the lens is:

$A$ thin equiconvex lens is made of glass of refractive index $1.5$ and its focal length is $0.2 \ m$. If it acts as a concave lens of $0.5 \ m$ focal length when dipped in a liquid,the refractive index of the liquid is:

$A$ concave lens of glass,refractive index $1.5$,has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

$A$ convex lens of focal length $f$ produces a real image $n$ times the size of the object. The image distance is

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is