Question

What happens to the focal length of convex lens when it is immersed in water? Refractive index of the material of lens is greater than that of water.

Answer 1

We can find the answer from lens maker’s formula.

Let $n_1$be absolute refractive index of material of glass and

$n_2$ be absolute refractive index of water.

The refractive index of thin lens in air is given by

$\frac{1}{f_a}$=$\frac{n_1-1}{\frac{1}{R_1}–\frac{1}{R_2}}$ ..................(1)

where $R_1$ and$R_2$ are radii of curvature of two surfaces of the convex lens.

Now, when lens is kept in water ,

we have to use refractive index of material of lens relative to water.

This is $\frac{n1}{n2}$

So,replacing in the above expression $n_1$ by$\frac{n_1}{n_2}$, we get

$\frac{1}{f_w}$=($\frac{n1}{n2}-1$)($\frac{1}{R_1}-\frac{1}{R_2}$)………………(2)

Comparison of Eq.(1) with Eq.(2) shows that$f_w>f_a$

Focal length increases.