Question

The defective eye of a person has near point 0.5 m and distant point 3 m. The power for corrective lens required for reading is :

• A. +3 D
• B. -3 D
• C. +1/3 D
• D. +1 D

Hint:

For reading, the object is placed at a near distance, so we correct for hyperopia using a convex (positive) lens.
This immediately eliminates negative options like (B).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
A person has a defective eye with a near point of \(0.5\text{ m}\) (instead of the normal near point) and a far point of \(3\text{ m}\). We need to determine the power of the lens required for reading.

Step 2: Key Formula or Approach:
Reading requires near vision correction.
We use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] and the power formula:
\[ P = \frac{1}{f\text{ (in meters)}} \]

Step 3: Detailed Explanation:

• The person wants to read a book.

• If we assume a standard comfortable reading distance (object distance \(u\)) of \(33.3\text{ cm}\) (or \(-1/3\text{ m}\)):
- Object distance (\(u\)) = \(-1/3\text{ m}\)
- To see the text clearly, the corrective lens must form a virtual image of this book at the person's actual near point, which is \(0.5\text{ m}\).
- Image distance (\(v\)) = \(-0.5\text{ m} = -1/2\text{ m}\)

• Using the lens formula in terms of power (\(P = 1/f\)):
\[ P = \frac{1}{v} - \frac{1}{u} \] \[ P = \frac{1}{-1/2} - \frac{1}{-1/3} \] \[ P = -2 + 3 = +1\text{ D} \]

• (Note: If the standard near point of a normal eye is assumed to be \(25\text{ cm}\), then \(u = -0.25\text{ m}\), which gives \(P = -2 + 4 = +2\text{ D}\). However, based on the standard question options where \(+2\text{ D}\) is not available, the question assumes a reading distance of \(33.3\text{ cm}\) to yield \(+1\text{ D}\)).


Step 4: Final Answer:
The power of the corrective lens required for reading is +1 D.