Two identical symmetric double convex lenses of focal length f are cut into two equal parts $$L_{1}, L_{2}$$ by AB plane and $$L_{3}, L_{4}$$ by XY plane as shown in figure respectively. The ratio of focal lengths of lenses $$L_{1}$$ and $$L_{3}$$ is

For the original equiconvex lens of focal length $$f$$:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = \frac{2(\mu - 1)}{R}$$

For part $$L_1$$ (cut along horizontal plane AB):

The curvature of the surfaces remains unchanged. $$f_1 = f$$

For part $$L_3$$ (cut along vertical plane XY):

One surface becomes plane ($$R_2 = \infty$$), making it a planoconvex lens.

$$\frac{1}{f_3} = (\mu - 1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = \frac{\mu - 1}{R} = \frac{1}{2f} \implies f_3 = 2f$$

Ratio of focal lengths of $$L_1$$ and $$L_3$$: $$\frac{f_1}{f_3} = \frac{f}{2f} = \frac{1}{2}$$