The figure shows an equiconvex lens of focal | vedclass

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(A) According to the Lens Maker's Formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \ri

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$2 f$

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(A) According to the Lens Maker's Formula,the focal length $f$ of a lens is given by $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$.For an equiconvex lens,$R_1 = R$ and $R_2 = -R$,so $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} ight) = \frac{2(\mu - 1)}{R}$.When the lens is cut along the axis $AB$ (perpendicular to the principal axis),the radius of curvature of the surfaces remains the same for each half.For each half,the lens becomes a plano-convex lens where one surface has radius $R$ and the other is flat (infinite radius,$R' = \infty$).Applying the formula for one half: $\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} ight) = \frac{\mu - 1}{R}$.Comparing this with the original focal length,we see that $\frac{1}{f'} = \frac{1}{2} 	imes \frac{2(\mu - 1)}{R} = \frac{1}{2} 	imes \frac{1}{f}$.Therefore,$f' = 2f$.

The figure shows an equiconvex lens of focal length $f$. If the lens is cut along $AB$,the focal length of each half will be

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