Question

An equiconvex lens of focal length F is cut in to two equal parts along the vertical axis. The focal length of each part will be

• A. \(4F\)
• B. \(\frac{F}{2}\)
• C. \(2F\)
• D. \(F\)

Hint:

Each half is plano-convex, with half the power.

Answer 1

The Correct Option is C

Step 1: Combine the halves:
Two identical thin lenses in contact have power $P=P_1+P_2$. The two halves put together give the original lens, so $\dfrac1F=\dfrac1f+\dfrac1f$.

Step 2: Solve:
$\dfrac1F=\dfrac2f$, so $f=2F$.

Step 3: Option:
(C).

Final Answer:
Two equal halves in contact give back the lens with F = f/2. \[ \boxed{C} \]