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(D) From the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens with equal radii of curvatu

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$2 f, \frac{P}{2}$

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(D) From the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For a convex lens with equal radii of curvature $R$,we have $R_1 = R$ and $R_2 = -R$. Thus,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$.So,$\frac{1}{f} = \frac{2(\mu - 1)}{R}$.When one surface is made plane by grinding,the new radii are $R_1 = R$ and $R_2 = \infty$.The new focal length $f'$ is given by $\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$.Comparing the two equations,we see that $\frac{1}{f'} = \frac{1}{2} \left( \frac{1}{f} \right)$,which implies $f' = 2f$.Since focal power $P = \frac{1}{f}$,the new power $P' = \frac{1}{f'} = \frac{1}{2f} = \frac{P}{2}$.Therefore,the new focal length is $2f$ and the new power is $\frac{P}{2}$.

The radii of curvature of both the surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are:

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