If the aperture of a lens is halved,then the image will be:

$A$ 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:

$A$ real object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between the object and its image will be

$A$ thin convex lens is made of a material of refractive index $1.6$. An object is kept at a distance of $u$ from the lens on the principal axis as shown in the figure. The radii of curvature of the surfaces are $10 \, cm$ and $5 \, cm$. Now,the lens is reversed such that the face having radius of curvature $5 \, cm$ lies close to the object. The difference in image position as obtained for both the cases is equal to ......$u$.

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

$A$ convex lens has a power of $+5.0 \, D$ in air,where the refractive index of the lens material is $_a\mu_g = 1.5$. In a liquid of what refractive index should it be immersed so that it acts as a concave lens of focal length $100 \, cm$?