Question

A double convex lens, made from a material of refractive index ${\mu }_1$, is immersed in a liquid of refractive index ${\mu }_2$ where ${\mu }_2>{\mu }_1$. What change, if any, would occur in the nature of the lens?

Answer 1

The nature of the lens will not change; shape, size, density etc will remain the same.
Now, the focal length will change.
The lens maker's formula is:

$\frac{1}{f}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

where f is the focal length, n2 the refractive index of the lens, n1 the refractive index of the medium and R1 and R2 the curvature radius of each lens's face.

In the air where n1=1 < n2, for a particular lens we will have:$\frac{1}{f}=(\frac{{\mu }_1}{1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

In a medium where n1> n2, for this particular lens we will have:$\dfrac{1}{f}=(\dfrac{\mu_1}{\mu_2}-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$