A convex lens has power $P$ . It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in the figure) then incorrect option

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Power of $L_{2} = \frac{P}{2}$

Power of $L_{1} = \frac{P}{2}$

Power of $L_{3} = \frac{P}{2}$

Power of $L_{1} = P$

answer is A.

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Detailed Solution

The power when cut along principal axis

$\begin{array}{l} R_{1} = + R, R_{2} = \infty \\ ∴ \frac{1}{f^{''}} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ = (μ - 1) (\frac{1}{R} - \frac{1}{\infty}) = (\frac{μ - 1}{R}) = \frac{1}{2 f} \Rightarrow P_{2} = P_{3} = \frac{P}{2} \end{array}$

The power if the lens is cut perpendicular to the principle axis

$\begin{array}{l} R_{1} = + R, R_{2} = - R \\ ∴ \frac{1}{f} = (μ - 1) [\frac{1}{R} - (\frac{1}{- R})] \Rightarrow \frac{1}{f} = (μ - 1) (\frac{1}{R} + \frac{1}{R}) \\ = (μ - 1) \frac{2}{R} = \frac{1}{f^{'}} \Rightarrow P_{1} = P \end{array}$