A person wears glases of power -2.5 D. The defect of the eye and far point of the person without glasses are respectively

Options

A. farsightness 40 cm
B. nearsightness 40 cm
C. nearsightness 250 cm
D. astigmatism 40 cm

Answer

To solve the problem, we need to determine the defect of the eye and the far point of a person who wears glasses with a power of -2.5 D.

### Step-by-Step Solution:

1. **Understanding the Power of Glasses**:
The power (P) of the glasses is given as -2.5 D. The negative sign indicates that the glasses are concave lenses, which are used to correct nearsightedness (myopia).

**Hint**: Remember that a negative power indicates a concave lens, which is used for nearsightedness.

2. **Calculating the Focal Length**:
The relationship between power (P) and focal length (f) is given by the formula:
\[ P = \frac{1}{f} \]
Rearranging this gives:
\[ f = \frac{1}{P} \]
Substituting the value of P:
\[ f = \frac{1}{-2.5} = -0.4 \text{ meters} = -40 \text{ cm} \]

**Hint**: Converting meters to centimeters can help in understanding the distance better.

3. **Understanding the Defect of the Eye**:
Since the person is nearsighted, their far point (the maximum distance at which they can see clearly) is at a distance less than infinity. The focal length we calculated (-40 cm) indicates that the far point of the person without glasses is at 40 cm.

**Hint**: For nearsighted individuals, the far point is closer than normal vision, which is typically at infinity.

4. **Using the Lens Maker's Formula**:
We can use the lens maker's formula to confirm the far point:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Here, for a nearsighted person, the object (far point) is at a distance (u) of -40 cm (since it is a virtual object for the lens), and the image (v) is at infinity (which is considered as +∞).
Substituting the values:
\[ \frac{1}{-40} = \frac{1}{\infty} - \frac{1}{-40} \]
This confirms that the far point is indeed at 40 cm.

**Hint**: Remember that in lens formulas, distances measured in the direction of the incoming light are negative.

### Conclusion:
The defect of the eye is nearsightedness (myopia), and the far point of the person without glasses is 40 cm.

### Final Answer:
- **Defect of the eye**: Nearsightedness (Myopia)
- **Far point without glasses**: 40 cm