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It is useful to design lenses of derived focal length using surface of suitable radii of curvature. Lens maker formula helps in finding various unknowns and is widely applied.

Find the focal length of an equi convex lens whose each face has radius of curvature 12 cm. Given refractive index of glass is $\frac{3}{2}$.

12 cm

The correct answer is option (4) : 12 cm

From lens maker's formula

$\frac{1}{f}=(\mu -1) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=\left(\frac{3}{2}-1\right)\left[\frac{1}{12}-\left(-\frac{1}{12}\right)\right]=\frac{1}{2}\times \frac{2}{12}$

$⇒f= 12 \, cm $