Correct Option: (B) P

Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]

The power of a thin lens is given by the Lens Maker's Formula:

$$P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

For a double-convex lens with equal radii of curvature ($R_1 = R$ and $R_2 = -R$), the initial power $P$ is:

$$P = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) = (\mu - 1) \left( \frac{2}{R} \right)$$

Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]

When the lens is cut perpendicular to the principal axis, the thickness of the lens is halved, but the curvature of the faces remains unchanged. For one of the resulting plano-convex lenses:

• The first surface is convex with radius $R_1 = R$.
• The second surface is flat (plane), so $R_2 = \infty$.

The power of the new lens ($P'$) is:

$$P' = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right)$$ $$P' = (\mu - 1) \left( \frac{1}{R} - 0 \right) = \frac{(\mu - 1)}{R}$$

Step 3 (Final Concluded Answer): [1/3 Mark]

Comparing the expression for $P'$ with the original power $P = \frac{2(\mu - 1)}{R}$:

$$P' = \frac{1}{2} \left[ \frac{2(\mu - 1)}{R} \right] = \frac{P}{2}$$

Note: While the calculation shows $P/2$, in the context of this specific board question, if the lens is cut perpendicular to the principal axis, the focal length doubles and power halves. Therefore, the correct answer is (D) P/2.

💡 EL10 Concept Insight: Imagine the lens is like a magnifying glass; when you slice it in half like a loaf of bread, you are making the lens thinner, which makes it less powerful at bending light, so the power drops to exactly half of what it was before.