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(C) For a symmetric convex lens,the radii of curvature are $R_1 = R$ and $R_2 = -R$.Using the Lens Maker's formula:$\frac{1}{f} = (\mu-1) \left[ \frac

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$2 f$

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(C) For a symmetric convex lens,the radii of curvature are $R_1 = R$ and $R_2 = -R$.Using the Lens Maker's formula:$\frac{1}{f} = (\mu-1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$$\frac{1}{f} = (\mu-1) \left[ \frac{1}{R} - \frac{1}{-R} \right] = (\mu-1) \left[ \frac{2}{R} \right]$Thus,$R = 2f(\mu-1)$.When the lens is cut vertically into two identical plano-convex lenses,for one such lens,the new radii of curvature are $R_1' = R$ and $R_2' = \infty$.Let the focal length of the plano-convex lens be $f'$. Using the Lens Maker's formula again:$\frac{1}{f'} = (\mu-1) \left[ \frac{1}{R_1'} - \frac{1}{R_2'} \right]$$\frac{1}{f'} = (\mu-1) \left[ \frac{1}{R} - \frac{1}{\infty} \right] = \frac{\mu-1}{R}$Substituting $R = 2f(\mu-1)$:$\frac{1}{f'} = \frac{\mu-1}{2f(\mu-1)} = \frac{1}{2f}$Therefore,$f' = 2f$.

$A$ convex lens has its radii of curvature equal. The focal length of the lens is $f$. If it is divided vertically into two identical plano-convex lenses by cutting it,then the focal length of the plano-convex lens is ($\mu$ = the refractive index of the material of the lens).

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