$A$ thin lens is made of a material with a refractive index of $\mu = 1.5$. It is a biconvex lens. When it is immersed in water,it behaves as a/an:

Explain how the image is formed by a thin convex lens and derive the lens maker's formula: $\frac{1}{f} = (n_{21} - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.

The distance between an object and its image produced by a converging lens is $0.72 \ m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04 \ m$ towards the lens?

$A$ lens is made of glass having an index of refraction $1.5$. One side of the lens is flat and the other side is convex with a radius $R$. If an object is placed $60 \ cm$ towards the convex side of the lens, the image is formed at $120 \ cm$ on the other side of the lens. The value of $R$ is (in $cm$)

Where should an object be placed on the axis of a convex lens of focal length $8 \ cm$ so as to achieve magnification of $-4$ (in $cm$)? (Distances are measured from the optic centre of the lens)

At what distance from a biconvex lens of focal length $F$ must an object be placed so that the distance between the object and its real image is minimal?