Q.7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Given the near point for the hypermetropic eye is 1m.
Assume the near point of the normal eye is 25 cm.
The object distance will be, (Normal near point).
The image distance, (Near point of this defective eye) or .
Then the focal length can be found from the lens formula:
Substituting the values in the equation, we obtain
Hence the power of the lens will be:
Thus, the power of the convex lens required will be