The focal length of a thin biconvex lens is $20 \ cm$. When an object is moved from a distance of $25 \ cm$ in front of it to $50 \ cm$,the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is

Where should an object be placed on the axis of a convex lens of focal length $8 \ cm$ so as to achieve magnification of $-4$ (in $cm$)? (Distances are measured from the optic centre of the lens)

$A$ thin lens with focal length $f$ is to be used as a magnifying glass. Which of the following statements regarding the situation is true?

$A$ lens of refractive index '$\mu$' has focal length '$f$'. When the lens is immersed in a liquid of refractive index '$\mu_0$',its focal length becomes '$f_0$'. Then '$f_0$' is given by

$A$ glass convex lens $(\mu_g = 1.5)$ has a focal length of $8\, cm$ in air. What will be its focal length when it is immersed in water (in $, cm$)? $(\mu_w = 1.33)$

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 $cm$)