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:  \[\frac{1}{f} = (\mu - 1)\left[ \frac{1}{R_1} - \frac{1}{R_2} \right]\]

\[\frac{1}{f} = 0 . 6\left[ \frac{1}{20} - \frac{1}{30} \right]\]
\[f = \frac{0 . 6}{1060 \times 10}\]
 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
\[\frac{1}{f} = \left[ \frac{\mu_{lens}}{\mu_{water}} - 1 \right]\left[ \frac{1}{R_1} - \frac{1}{R_2} \right]\]
\[= \left( \frac{1 . 60}{1 . 33} - 1 \right)\left[ \frac{1}{60} \right]\]
\[= \frac{28}{133 \times 60} \simeq \frac{1}{300}\]
\[\Rightarrow\] f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.