Step 1: Understanding the Concept:
The focal length (\(f\)) of a thin lens is determined by its refractive index (\(\mu\)) and the radii of curvature (\(R_1, R_2\)) of its two refracting surfaces. Cutting a lens modifies its physical dimensions, which can either leave its focal length unchanged or transform its geometry entirely depending on the orientation of the cut.
Step 2: Key Formula or Approach:
We utilize the Lens Maker's Formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an uncut, symmetric double convex lens with radii of curvature magnitude \(R\), the surfaces are defined by \(R_1 = R\) and \(R_2 = -R\):
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right) = \frac{2(\mu - 1)}{R} \implies f = \frac{R}{2(\mu - 1)} \]
Step 3: Detailed Explanation:
Analyzing \(L_1\) (Horizontal cut through plane AB):
The plane \(AB\) passes horizontally right along the principal axis. This splits the lens into an upper half \(L_1\) and a lower half \(L_2\).
The curvature of both the front surface (\(R_1 = R\)) and back surface (\(R_2 = -R\)) is completely preserved.
Only the aperture (the physical surface area available to capture light) is halved, reducing the intensity of the image formed.
Because the radii of curvature do not change, the focal length remains exactly equal to the original value:
\[ f_{L_1} = f \]
Analyzing \(L_3\) (Vertical cut through plane XY):
The plane \(XY\) cuts the lens vertically down the middle, perpendicular to the principal axis. This divides the symmetric double convex lens into two identical plano-convex pieces, \(L_3\) and \(L_4\).
For the part \(L_3\), the front surface remains curved (\(R_1 = R\)).
The rear cut surface is now a perfectly flat plane, which corresponds to an infinite radius of curvature (\(R_2 = \infty\)).
Substituting these boundary properties into the Lens Maker's Formula:
\[ \frac{1}{f_{L_3}} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R} \]
Comparing this equation to the original lens expression (\(\frac{1}{f} = \frac{2(\mu - 1)}{R}\)), we can substitute to find:
\[ \frac{1}{f_{L_3}} = \frac{1}{2} \times \frac{1}{f} \implies f_{L_3} = 2f \]
Thus, cutting the lens vertically halves its optical bending power, which doubles its focal length.
Step 4: Calculating the Ratio:
We evaluate the ratio of the focal length of lens \(L_1\) to lens \(L_3\):
\[ \text{Ratio} = \frac{f_{L_1}}{f_{L_3}} = \frac{f}{2f} = \frac{1}{2} \]
Final Answer:
The ratio of focal lengths of lenses \(L_1\) and \(L_3\) is 1 : 2.