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: Understand what the corrective lens must achieve.
The person cannot focus on anything closer than $0.5\text{ m}$, but wants to read at a comfortable distance of about $\dfrac{1}{3}\text{ m}$, so the lens must take an object at $\dfrac{1}{3}\text{ m}$ and throw its image out to $0.5\text{ m}$, right where the eye can actually focus.
Step 2: Assign the object and image distances. \[ u = -\frac{1}{3}\text{ m}, \qquad v = -0.5\text{ m} \] both taken negative since both the book and its virtual image lie on the same side as the incoming light.
Step 3: Solve for the focal length needed. \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-0.5} - \frac{1}{-1/3} = -2 + 3 = 1 \implies f = 1\text{ m} \]
Step 4: Convert focal length to power. \[ P = \frac{1}{f} = \frac{1}{1} = +1\text{ D} \]
\[ \boxed{+1\text{ D}} \]