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 . Choose the incorrect option for the reported pieces.

Question

What is the focal length of a lens if its power is +2 D?

Answer 1

To solve this problem, let's analyze how the power of a lens changes when it is cut.

When a convex lens with power \(P\) is cut along its principal axis, each half lens will have the same focal length but different areas. The power of a lens based on its geometry depends on its material and curvature; when a lens is bisected along its principal axis, each half lens retains half the power of the original lens. Thus, both pieces have power \( \frac{P}{2} \), maintaining the same focal length as the original full lens.
Now, one of these halves, e.g., \(L_1\), is cut again into two parts, but this time perpendicular to the principal axis. This means each piece now becomes a smaller lens covering half the area of the original lens but due to perpendicular cutting, there is no change in curvature or material which determines the lens power, so each keeps the same power it had before, i.e., \( \frac{P}{2} \).
Since we have cut \(L_1\) along the surface that influences power distribution (i.e., perpendicular to the axis), the lenses do not change their power due to this symmetry.

Let's analyze the given options in light of these details:

Option 1: Power of \(L_1 = \frac{P}{2}\) - This is correct as when divided vertically across the principal axis, the power remains \( \frac{P}{2} \) for the half pieces individually.
Option 2: Power of \(L_2 = \frac{P}{2}\) - This is also correct as similar reasoning applies here as \(L_1\).
Option 3: Power of \(L_3 = \frac{P}{2}\) - Similarly correct since the perpendicular division doesn't change the power.
Option 4: Power of \(L_1 = P\) - Incorrect because upon splitting the original lens, \(L_1\) can only have half of the initial power, i.e., \(\frac{P}{2}\), not the full power.

Therefore, the incorrect option is: \(L_1 = \frac{P}{2}\), where the power is stated to be \(P\), which contradicts the result of the cut.

Answer 2

To solve this problem, let's analyze how the power of a lens changes when it is cut.

When a convex lens with power \(P\) is cut along its principal axis, each half lens will have the same focal length but different areas. The power of a lens based on its geometry depends on its material and curvature; when a lens is bisected along its principal axis, each half lens retains half the power of the original lens. Thus, both pieces have power \( \frac{P}{2} \), maintaining the same focal length as the original full lens.
Now, one of these halves, e.g., \(L_1\), is cut again into two parts, but this time perpendicular to the principal axis. This means each piece now becomes a smaller lens covering half the area of the original lens but due to perpendicular cutting, there is no change in curvature or material which determines the lens power, so each keeps the same power it had before, i.e., \( \frac{P}{2} \).
Since we have cut \(L_1\) along the surface that influences power distribution (i.e., perpendicular to the axis), the lenses do not change their power due to this symmetry.

Let's analyze the given options in light of these details:

Option 1: Power of \(L_1 = \frac{P}{2}\) - This is correct as when divided vertically across the principal axis, the power remains \( \frac{P}{2} \) for the half pieces individually.
Option 2: Power of \(L_2 = \frac{P}{2}\) - This is also correct as similar reasoning applies here as \(L_1\).
Option 3: Power of \(L_3 = \frac{P}{2}\) - Similarly correct since the perpendicular division doesn't change the power.
Option 4: Power of \(L_1 = P\) - Incorrect because upon splitting the original lens, \(L_1\) can only have half of the initial power, i.e., \(\frac{P}{2}\), not the full power.

Therefore, the incorrect option is: \(L_1 = \frac{P}{2}\), where the power is stated to be \(P\), which contradicts the result of the cut.

The Correct Option is A