Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is `1 metre`. What is the power of the lens required to correct this defect ? Assume that the near point of the normal eye is `25 cm`.

Explanation

To solve the problem of correcting hypermetropia and finding the power of the lens required, we will follow these steps:

### Step 1: Understand Hypermetropia
Hypermetropia, or farsightedness, is a condition where distant objects can be seen clearly, but close objects appear blurry. In a hypermetropic eye, the image of a near object is formed behind the retina.

### Step 2: Identify Given Data
- Near point of hypermetropic eye (D) = 1 meter = 100 cm
- Near point of normal eye (N) = 25 cm
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