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(C) For a biconvex lens with equal radii of curvature $R_1 = R$ and $R_2 = -R$, the Lens Maker's formula is given by:$\frac{1}{f} = (\mu - 1) \left[ \

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$2 f$

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(C) For a biconvex lens with equal radii of curvature $R_1 = R$ and $R_2 = -R$, the Lens Maker's formula is given by:$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$Substituting the values:$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{-R} \right] = (\mu - 1) \left[ \frac{2}{R} \right]$Thus, $R = 2f(\mu - 1)$.When the lens is cut vertically into two identical plano-convex lenses, each new lens has one surface with radius $R$ and the other surface as a plane (radius $\infty$).Using the Lens Maker's formula for the new lens with focal length $f'$:$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} - \frac{1}{\infty} \right]$$\frac{1}{f'} = (\mu - 1) \left[ \frac{1}{R} \right]$$f' = \frac{R}{\mu - 1}$Substituting $R = 2f(\mu - 1)$:$f' = \frac{2f(\mu - 1)}{\mu - 1} = 2f$.

$A$ 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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