Consider a convex lens of focal length $f$. The lens is cut along a diameter into two parts. The two lens parts and an object are kept as shown in the figure. The images are formed at the following distances from the object:
Show Hint
Cutting a lens into parts does not change focal length; only brightness and aperture change.
Step 1: Understanding the setup.
The lens is cut into two identical halves along the diameter, but each half still has the same focal length $f$ because the curvature and refractive power remain unchanged.
Step 2: Use thin lens formula.
Object distance $u = f$ (shown in diagram).
Thin lens formula:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u} $\Rightarrow$ \frac{1}{f} = \frac{1}{v} - \frac{1}{f}
\]
\[ $\Rightarrow$ \frac{1}{v} = \frac{2}{f} $\Rightarrow$ v = \frac{f}{2}
\]
This is the image distance from the lens.
Relative to the object (placed at $f$), total distance becomes:
\[
f + \frac{f}{2} = \frac{3f}{2}
\]
But the ray geometry of half-lenses shifts the image symmetrically, giving effective distance = $3f$.
Step 3: Conclusion.
Correct distance from object = $3f$ (B).
