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).

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Solution

## 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, (μwater) = 1.33 (a) When the lens is placed in air, Using lens maker formula: $\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$ $\frac{1}{f}=0.6\left[\frac{1}{20}-\frac{1}{30}\right]\phantom{\rule{0ex}{0ex}}$ $f=\frac{0.6}{1060×10}$ 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 $\frac{1}{f}=\left[\frac{{\mathrm{\mu }}_{\mathrm{lens}}}{{\mathrm{\mu }}_{\mathrm{water}}}-1\right]\left[\frac{\mathit{1}}{{R}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{R}_{\mathit{2}}}\right]$ $=\left(\frac{1.60}{1.33}-1\right)\left[\frac{1}{60}\right]$ $=\frac{28}{133×60}\simeq \frac{1}{300}$ $⇒$f = 300 cm. Thus, the focal length of the lens is 300 cm when it is placed in water.

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