# What will happen to the focal length of convex lens which is placed inside water ?

Feb 2, 2017

I would think that it gets bigger.

#### Explanation:

We know from the famous Lensmaker´s Formula that:
$\frac{1}{f} = \frac{{n}_{l} - {n}_{m}}{n} _ m \left(\frac{1}{R} _ 1 + \frac{1}{R} _ 2\right)$
where:
$f =$ focal length;

${n}_{l} =$ index of refraction of lens (glass);

${n}_{m} =$ index od refraction of the medim (in which the lens is placed);

${R}_{1} \mathmr{and} {R}_{2}$ radii of curvature of lens surfaces.

Consider an exampe:

A lens that has a focal length of $+ 20 c m$ has equal radii, so that ${R}_{1} = {R}_{2} = R$, and is made of glass of ${n}_{l} = 1.6$; you place it into water (${n}_{m} = 1.33$). Remember that we can consider the index of air as $1$. We have:

In air:
$\frac{1}{20} = \frac{1.6 - 1}{1} \left(\frac{2}{R}\right)$

In water:
$\frac{1}{f '} = \frac{1.6 - 1.33}{1.33} \left(\frac{2}{R}\right)$
substitute one into the other for $\left(\frac{2}{R}\right)$:
$\frac{1}{20 \left(1.6 - 1\right)} = \frac{1.33}{f ' \left(1.6 - 1.33\right)}$
rearranging:
$f ' = 1.33 \frac{20 \left(1.6 - 1\right)}{1.6 - 1.33} \approx 60 c m$