A biconvex lens has radii of curvature 20 cm each. If the refractive index of the material of the lens is 1.5, what is its focal length?

20 cm

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-10 cm

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10 cm

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-20 cm

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Solution

The correct option is A 20 cm
In a biconvex lens, centre of curvature of the first surface is on the positive side of the lens and that of the second surface is on the negative side.
Thus, $R_{1}$ = 20 cm and $R_{2}$ = -20 cm.
We have,
$\frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) o r, \frac{1}{f} = (1.5 - 1) (\frac{1}{20 c m} - \frac{1}{- 20 c m}) o r, f = 20 c m$

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