The ratio of the focal lengths of a convex lens when kept in air and when it is immersed in a liquid is $1: 2$. If the refractive index of the material of the lens is $1.5$,then the refractive index of the liquid is

An object has an image thrice of its original size when kept at $8 \ cm$ and $16 \ cm$ from a convex lens. The focal length of the lens is:

The distance between an object and the screen is $80 \, cm$. $A$ lens produces an image on the screen when the lens is placed at either of the positions $20 \, cm$ apart. The focal length of the lens is ....... $cm$.

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?

$A$ 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?

Will the focal length of a lens for red light be more,same,or less than that for blue light?