Discussion
A convex lens having power P is cut into two halves perpendicular to the principal axis. Then the power of each piece is
Answer: B
The power of each piece is \(\frac{P}{2}\).
Cutting a biconvex lens perpendicular to its principal axis splits it into two plano-convex lenses. For an equiconvex lens of radius R, \(\frac{1}{f} = (\mu - 1)\frac{2}{R}\); for each plano-convex half, \(\frac{1}{f'} = (\mu - 1)\frac{1}{R}\). So \(f' = 2f\) and the power halves.
A cut along the axis (parallel to it) would leave the focal length, and so the power P, unchanged; that is the trap in option A. 2P and 4P go the wrong way.
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