In the displacement method,the distance between the object and the screen is $70 \; cm$ and the focal length of the lens is $16 \; cm$. Find the separation between the magnified and diminished image positions of the lens. (in $; cm$)

The power of a thin convex lens placed in air is $+4 \ D$. The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then

$A$ 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 $4/3$.

$A$ convex lens made of glass has a focal length of $0.15 \,m$ in air. If the refractive index of glass is $\frac{3}{2}$ and that of water is $\frac{4}{3}$, what is the focal length of the lens when immersed in water (in $\,m$)?

$A$ slanted object $AB$ is placed on one side of a convex lens as shown in the diagram. The image is formed on the opposite side. The angle made by the image with the principal axis is

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