$A$ lens having refractive index $1.6$ has a focal length of $12 \ cm$ when it is in air. Find the focal length of the lens when it is placed in water. (Take the refractive index of water as $1.28$) (in $mm$)

$A$ thin convex lens is placed just above an empty vessel of depth $80 \,cm$. The image of a coin kept at the bottom of the vessel is thus formed $20 \,cm$ above the lens. If now water is poured in the vessel up to a height of $64 \,cm$, what will be the approximate new position of the image? Assume that the refractive index of water is $4/3$.

What are paraxial rays?

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 ...... $cm$.

$A$ concave lens made of material of refractive index $1.6$ is immersed in a medium of refractive index $2.0$. The two surfaces of the concave lens have the same radius of curvature $0.2 \, m$. The lens will behave as a

$A$ thin converging lens of focal length $f=25 \,cm$ forms the image of an object on a screen placed at a distance of $75 \,cm$ from the lens. The screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object has to be shifted, so that its image on the screen is sharp again, is (in $\,cm$)