The near point of hypermetropic eye is 40 cm . What is the power of the lens used to correct this defect ?

Options

A. 2.5 D
B. 1.5 D
C. `-1.5 D `
D. 0.5 D

Explanation

To solve the problem of finding the power of the lens used to correct the hypermetropic eye, we can follow these steps:

### Step 1: Understand the Problem
A hypermetropic eye has a near point that is farther than the normal near point. The near point for a normal eye is typically 25 cm. In this case, the near point of the hypermetropic eye is given as 40 cm.

### Step 2: Determine the Focal Length of the Lens
To correct the hypermetropic eye, we need to bring the near point from 40 cm to the normal near point of 25 cm. We can use the lens formula:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance (which will be the near point of the normal eye, -25 cm),
- \( u \) is the object distance (which will be the near point of the hypermetropic eye, -40 cm).

### Step 3: Substitute the Values into the Lens Formula
Substituting the values into the lens formula:
\[ \frac{1}{f} = \frac{1}{-25} - \frac{1}{-40} \]

### Step 4: Calculate the Focal Length
Now, let's calculate:
\[ \frac{1}{f} = -\frac{1}{25} + \frac{1}{40} \]

Finding a common denominator (which is 200):
\[ \frac{1}{f} = -\frac{8}{200} + \frac{5}{200} = -\frac{3}{200} \]

Thus,
\[ f = -\frac{200}{3} \approx -66.67 \text{ cm} \]

### Step 5: Calculate the Power of the Lens
The power \( P \) of the lens is given by the formula:
\[ P = \frac{1}{f \text{ (in meters)}} \]
Converting focal length from cm to meters:
\[ f = -0.6667 \text{ m} \]
Now substituting into the power formula:
\[ P = \frac{1}{-0.6667} \approx -1.5 \text{ diopters} \]

### Conclusion
The power of the lens used to correct the hypermetropic eye is approximately -1.5 diopters.

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