Question

Assuming the person wears contacts, what power of contact lens must be used to correct the vision of a nearsighted person whose far point is 1.50 m?

Answer 1

Since the nearsighted eye over converges light rays, the correction for nearsightedness is to place a diverging spectacle lens in front of the eye. To determine the spectacle power needed for correction, you must know the person’s far point—that is, you must know the greatest distance at which the person can see clearly. Then the image produced by a spectacle lens must be at this distance or closer for the nearsighted person to be able to see it clearly. That means the spectacle lens must produce an image 1.5 m from the eye for an object very far away.

Thus, for object at ∞ , i.e. ,u= ∞ , and image at 1.5m , i.e. v=-1.5m (the image distance is negative since it is on the same side as the object )

The focal length of the corrective lens would be

1/f = 1/v + 1/u

1/f = [1/(-1.5)] + 1/∞

Thus, f=-1.5m

and so the power of corrective contact lens would be

P=1/f =- 0.66D