The near point of a patient's eye is 50.0 cm. What power (in diopters) must a corrective lens have to enable the eye to see clearly an object 25.0 cm away?

1. Analyze the Question

The patient is suffering from hypermetropia (long-sightedness). They cannot see objects clearly at the normal near point (25 cm) because their own near point has receded to 50 cm. We need to find the power of a lens that will take an object placed at 25 cm and create a virtual image at the patient's actual near point (50 cm), so the eye can focus on it.

2. Explain the Concept

To solve this, we use the Lens Formula:

Where:

• $f$ = focal length of the lens.
• $u$ = object distance (where we want to place the object = 25 cm).
• $v$ = image distance (where the eye can actually see = -50 cm). • Note: We use a negative sign for $v$ because the image formed by a corrective lens for a long-sighted eye must be virtual and on the same side as the object.

The Power ($P$) of the lens in diopters is given by:

3. Show the Solution

Step 1: Identify the values

• $u = 25\text{ cm} = 0.25\text{ m}$
• $v = -50\text{ cm} = -0.50\text{ m}$

Step 2: Calculate the focal length ($f$)

Step 3: Calculate the Power ($P$) Since $P = \frac{1}{f}$ and we already found $\frac{1}{f} = 2$:

4. Identify the Answer

The calculation yields +2.0 diopters. This matches option A.

5. Exam Tip

When dealing with eye defects in physics:

• Hypermetropia (Long-sightedness): Always requires a converging (convex) lens, which results in a positive (+) power.
• Myopia (Short-sightedness): Always requires a diverging (concave) lens, which results in a negative (-) power.
• Always ensure your distance units are in meters before calculating power, or convert your final result carefully!