A convex lens has a focal length f. It is cut | vedclass

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(D) According to the Lens Maker's Formula,the focal length $f$ of a thin convex lens with refractive index $\mu$ and radii of curvature $R_1 = R$ and

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$2f$

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(D) According to the Lens Maker's Formula,the focal length $f$ of a thin convex lens with refractive index $\mu$ and radii of curvature $R_1 = R$ and $R_2 = -R$ is given by:$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} ight) = (\mu - 1) \left( \frac{2}{R} ight)$So,$f = \frac{R}{2(\mu - 1)}$.When the lens is cut vertically along the dotted line,each part becomes a plano-convex lens.For a plano-convex lens,the radii of curvature are $R_1 = R$ and $R_2 = \infty$.Applying the Lens Maker's Formula for the new focal length $f'$:$\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} ight) = (\mu - 1) \left( \frac{1}{R} - 0 ight) = \frac{\mu - 1}{R}$.Therefore,$f' = \frac{R}{\mu - 1}$.Comparing $f'$ with $f$,we get $f' = 2 	imes \left( \frac{R}{2(\mu - 1)} ight) = 2f$.Thus,the focal length of each part will be $2f$.

$A$ convex lens has a focal length $f$. It is cut into two parts along the dotted line as shown in the figure. The focal length of each part will be

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