An equiconvex lens of focal length F is cut i | vedclass

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(C) The lens maker's formula is given by $\frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.For an equiconvex lens, $R_1 = R$ and

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$2F$

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(C) The lens maker's formula 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 $\frac{1}{F} = (\mu - 1) \left( \frac{2}{R} ight)$.When the lens is cut along the vertical axis, the radius of curvature of the new surfaces remains the same as the original surfaces ($R_1 = R$ and $R_2 = \infty$ for the new plano-convex lens).Applying the lens maker's formula for the new part: $\frac{1}{F'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} ight) = \frac{\mu - 1}{R}$.Comparing the two equations, we get $\frac{1}{F'} = \frac{1}{2} 	imes \frac{1}{F}$, which implies $F' = 2F$.

An equiconvex lens of focal length $F$ is cut into two equal parts along the vertical axis. The focal length of each part will be

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