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.

Answer 1

\[\text { Case I: Lens in air }\]

\[\text { Let the focal length of lens in air be F}_{air} \]

\[\text { Given that } \]

\[ F_{air} = 20 cm \]

\[ n_1 = 1 (air)\]

\[ n_2 = 1 . 6\]

\[\text { According to lens maker's formula }: \]

\[\frac{1}{F_{air}} = [\frac{n_2}{n_1} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]

\[\frac{1}{20} = [\frac{1 . 6}{1} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]

\[\frac{1}{20} = [0 . 6][\frac{1}{R_1} - \frac{1}{R_2}] . . . . . . (i)\]

\[\text { Case II: Lens in liquid }\]

\[\text { Let the focal length of lens in liquid be F}_{liquid} \]

\[\text { Given that } \]

\[ n_1 = 1 . 3 \text { (liquid) }\]

\[ n_2 = 1 . 6\]

\[\text { According to lens maker's formula }: \]

\[\frac{1}{F_{liquid}} = [\frac{1 . 6}{1 . 3} - 1][\frac{1}{R_1} - \frac{1}{R_2}]\]

\[\frac{1}{F_{liquid}} = [0 . 2307][\frac{1}{R_1} - \frac{1}{R_2}] . . . . . . (ii)\]

\[\text { Dividing (i) by (ii), we get}\]

\[\frac{F_{liquid}}{20} = \frac{0 . 6}{0 . 2307}\]

\[ F_{liquid} = 52 . 0156 cm\]