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Question
A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1·6. If it is immersed in a liquid of refractive index 1·3, find its new focal length.
Solution
\[\text { Case I: Lens in air }\]
\[\text { Let the focal length of lens in air be F}_{air} \]
\[\text { Given that } \]
\[ F_{air} = 20 cm \]
\[ n_1 = 1 (air)\]
\[ n_2 = 1 . 6\]
\[\text { According to lens maker's formula }: \]
\[\frac{1}{F_{air}} = [\frac{n_2}{n_1} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]
\[\frac{1}{20} = [\frac{1 . 6}{1} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]
\[\frac{1}{20} = [0 . 6][\frac{1}{R_1} - \frac{1}{R_2}] . . . . . . (i)\]
\[\text { Case II: Lens in liquid }\]
\[\text { Let the focal length of lens in liquid be F}_{liquid} \]
\[\text { Given that } \]
\[ n_1 = 1 . 3 \text { (liquid) }\]
\[ n_2 = 1 . 6\]
\[\text { According to lens maker's formula }: \]
\[\frac{1}{F_{liquid}} = [\frac{1 . 6}{1 . 3} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]
\[\frac{1}{F_{liquid}} = [0 . 2307][\frac{1}{R_1} - \frac{1}{R_2}] . . . . . . (ii)\]
\[\text { Dividing (i) by (ii), we get}\]
\[\frac{F_{liquid}}{20} = \frac{0 . 6}{0 . 2307}\]
\[ F_{liquid} = 52 . 0156 cm\]
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