A person uses a lens of power `+3D` to normalise vision. Near point of hypermetropic eye is

Options

A. `1m`
B. `1.66m`
C. `2m`
D. `0.66m`

Explanation

To find the near point of a hypermetropic eye using a lens of power +3D, we can follow these steps:

### Step 1: Understand the concept of power of a lens
The power \( P \) of a lens is given by the formula:
\[ P = \frac{1}{f} \]
where \( f \) is the focal length in meters. Given that the power of the lens is +3D, we can find the focal length.

### Step 2: Calculate the focal length
Using the formula for power:
\[ P = +3D \implies f = \frac{1}{P} = \frac{1}{3} \text{ meters} = \frac{100}{3} \text{ cm} \approx 33.33 \text{ cm} \]

### Step 3: Understand normal vision
Normal vision allows a person to see objects clearly at a distance of 25 cm. This means that for the hypermetropic eye, the lens must help focus light from objects at this distance.

### Step 4: Use the lens formula
The lens formula is:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where:
- \( f \) is the focal length of the lens (33.33 cm),
- \( v \) is the image distance (which we need to find),
- \( u \) is the object distance (which is -25 cm, as per the sign convention).

### Step 5: Substitute values into the lens formula
Substituting the known values into the lens formula:
\[ \frac{1}{\frac{100}{3}} = \frac{1}{v} - \frac{1}{-25} \]
This simplifies to:
\[ \frac{3}{100} = \frac{1}{v} + \frac{1}{25} \]

### Step 6: Solve for \( \frac{1}{v} \)
Rearranging gives:
\[ \frac{1}{v} = \frac{3}{100} - \frac{1}{25} \]
To combine the fractions, convert \( \frac{1}{25} \) to a fraction with a denominator of 100:
\[ \frac{1}{25} = \frac{4}{100} \]
Thus:
\[ \frac{1}{v} = \frac{3}{100} - \frac{4}{100} = -\frac{1}{100} \]

### Step 7: Find \( v \)
Taking the reciprocal gives:
\[ v = -100 \text{ cm} \]
This indicates that the near point of the hypermetropic eye is at a distance of 100 cm.

### Conclusion
The near point of the hypermetropic eye is 100 cm.

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