Question 180 of 295

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 figures). Choose the incorrect option for the reported pieces.

APower of ${L}_{1}=\dfrac{P}{2}$BPower of ${L}_{2}=\dfrac{P}{2}$CPower of ${L}_{3}=\dfrac{P}{2}$DPower of ${L}_{1}=P$

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Answer: (A) Power of ${L}_{1}=\dfrac{P}{2}$

Using Lens maker's formula, power is given by
$P=\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)$

When convex lens is cut into two parts, ${P}_{1}=\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{2}{R}\right)=P$

When half convex lens is cut into two parts,${P}_{2}=\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{1}{R}\right)=\dfrac{P}{2}$

Power of lens, ${P}_{3}=\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{1}{R}\right)=\dfrac{P}{2}$