The focal length of a converging lens in air is $R$. If it is dipped in water of refractive index $1.33$,then its focal length will be around (Refractive index of lens material is $1.5$).

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

$A$ glass convex lens has a refractive index of $1.55$ with both faces having the same radius of curvature. What is the radius of curvature required if the focal length is to be $20 \,cm$ (in $\,cm$)?

$A$ double convex thin lens made of glass (refractive index $\mu = 1.5$) has both radii of curvature of magnitude $20 \ cm$. Incident light rays parallel to the axis of the lens will converge at a distance $L$ such that $L = ...... \ cm$.

$A$ convex lens is placed between an object and a screen which are at a fixed distance $D$ apart. For one position of the lens,the magnification of the image formed on the screen is $m_1$. When the lens is shifted by a distance $d$,the magnification of the image formed on the same screen is $m_2$. The focal length of the lens is:

Assertion : Position of image approaches focus of a lens,only when object approaches infinity.
Reason : Paraxial rays incident parallel to principal axis intersect at the focus after refraction from lens.