JEE Physics 2025 Question with Solution

Given: Two identical symmetric double convex lenses each have focal length $f$. Lens parts $L_1, L_2$ are obtained by cutting one lens by plane $AB$, while $L_3, L_4$ are obtained by cutting the other lens by plane $XY$.

Find: The ratio of focal lengths of $L_1$ and $L_3$.

When a lens is cut by a plane passing through the principal axis as in the $AB$ cut, each part retains the same radii of curvature of the original lens. Therefore the focal length of $L_1$ remains unchanged:

$$f_1 = f$$

When the lens is cut by the $XY$ plane, the symmetric double convex lens behaves like a plano-convex part for each half, so its power becomes half of the original. Hence the new focal length becomes double:

$$P_3 = \frac{P}{2}$$

Since $P = \frac{1}{f}$,

$$\frac{1}{f_3} = \frac{1}{2f}$$

Therefore,

$$f_3 = 2f$$

Now,

$$\frac{f_1}{f_3} = \frac{f}{2f} = \frac{1}{2}$$

Therefore, the ratio of focal lengths of $L_1$ and $L_3$ is $1 : 2$. The correct option is D.

Given: Original focal length of each symmetric double convex lens is $f$.

Find: $f_1 : f_3$.

Use the fact that lens power is

$$P = \frac{1}{f}$$

For the lens cut by plane $AB$, the aperture is reduced but the curvatures are unchanged, so the focal length does not change:

$$f_1 = f$$

For the lens cut by plane $XY$, one curved surface of each half remains effective with half the original optical power. Hence,

$$P_3 = \frac{P}{2} = \frac{1}{2f}$$

This gives

$$f_3 = 2f$$

So,

$$f_1 : f_3 = f : 2f = 1 : 2$$

Therefore, the correct option is D.