Question

What happens to the focal length of a convex lens when it is immersed in water?

Answer 1

Step 1: Given and assume

Let,

The absolute refractive index of glass material be ${\mathrm{n}}_1$.

The absolute refractive index of water be ${\mathrm{n}}_2$.

Step 2: To find

We have to determine the focal length of a convex lens when it is immersed in water.

Step 3: Calculation

By using the lens maker's formula,

The refractive index of a thin lens in the air will be:

$\frac{1}{{\mathrm{f}}_{\mathrm{a}}}=\frac{{\mathrm{n}}_1-1}{{\displaystyle \frac{1}{{\mathrm{R}}_1}}-{\displaystyle \frac{1}{{\mathrm{R}}_2}}}$…(equation 1)

Here,

${\mathrm{R}}_1$ and ${\mathrm{R}}_2$ are the radii of curvature.

Now,

We have to use the refractive index of lens material relative to water that is $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$ when the lens is kept in water.

By replacing ${\mathrm{n}}_1$ by $\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}$, we get

$\Rightarrow \frac{1}{{\mathrm{f}}_{\mathrm{w}}}=\left(\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)$…(equation 2)

By comparing both the equation, we see

${\mathrm{f}}_{\mathrm{w}}>{\mathrm{f}}_{\mathrm{a}}$

Therefore, the focal length of a convex lens increases when it is immersed in water.