Two identical symmetric double convex lenses | vedclass

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(B) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When a len

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$1: 2$

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(B) The focal length of a lens is given by the lens maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When a lens is cut perpendicular to the principal axis (along the $AB$ plane),the radius of curvature of each surface remains the same,but the thickness is halved. However,the focal length of each part remains $f$. Thus,$f_{L_1} = f$.When a lens is cut parallel to the principal axis (along the $XY$ plane),the radius of curvature of one surface becomes infinite. The new focal length $f'$ is given by $\frac{1}{f'} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{(\mu - 1)}{R}$. Since $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = \frac{2(\mu - 1)}{R}$,we get $f' = 2f$. Thus,$f_{L_3} = 2f$.The ratio of the focal lengths is $\frac{f_{L_1}}{f_{L_3}} = \frac{f}{2f} = 1: 2$.

Two identical symmetric double convex lenses of focal length $f$ are cut into two equal parts $L_1, L_2$ by the $AB$ plane and $L_3, L_4$ by the $XY$ plane as shown in the figure respectively. The ratio of the focal lengths of lenses $L_1$ and $L_3$ is

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