Question

The radii of curvature of a lens are + 20 cm and + 30 cm. The material of the lens has a refracting index 1.6. Find the focal length of the lens (a) if it is placed in air, and (b) if it is placed in water (μ = 1.33).

Answer 1

Given,
Radii of curvature of a lens, (R₁₎ = +20 cm and R₂ = +30 cm
Refractive index of the material of the lens, (μ) = 1.6
Refractive index of water, (μ_water) = 1.33

(a)
When the lens is placed in air,
Using lens maker formula:

$\frac{1}{f}=(\mathrm{\mu }-1)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$
$\frac{1}{f}=0.6\left[\frac{1}{20}-\frac{1}{30}\right]$
$f=\frac{0.6}{1060\times 10}$
f = 100 cm
Thus, the focal length of the lens is 100 cm when it is placed in air.

(b)When the lens is placed in water

$\frac{1}{f}=\left[\frac{{\mathrm{\mu }}_{\mathrm{lens}}}{{\mathrm{\mu }}_{\mathrm{water}}}-1\right]\left[\frac{\mathit{1}}{R_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{R_{\mathit{2}}}\right]$
$=\left(\frac{1.60}{1.33}-1\right)\left[\frac{1}{60}\right]$
$=\frac{28}{133\times 60}\simeq \frac{1}{300}$
$\Rightarrow$f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.