A concavo-convex lens has an index of refraction 1.5 and the radii of curvature of its surface are 10 cm and 20 cm. The concave surface is upwards and it is filled with an oil of refractive index 1.6. Calculate the focal length of the oil-glass combination.

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Detailed Solution

For the glass lens, $\frac{1}{f} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}];$
Here $R_{1} = 10 cm, R_{2} = 20 cm and μ = 1 .5$
$\frac{1}{f_{1}} = (1 .5 - 1) [\frac{1}{10 cm} - \frac{1}{20 cm}] = \frac{1}{40 cm}$
For the oil lens, $\frac{1}{f_{2}} = (μ - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$
Here $R_{1} = 20 cm, R_{2} = \infty, and μ = 1 .6$
$\frac{1}{f} = (1 .6 - 1) [\frac{1}{20 cm} - \frac{1}{\infty}] = \frac{3}{100 cm}$
For the oil-glass combination $\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{40 cm} + \frac{3}{100 cm} = \frac{11}{200 cm} ∴ f = 18 .2 cm$

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