Question:medium

The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Updated On: Jan 13, 2026
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Solution and Explanation

The individual has myopia, an eye defect where the image forms in front of the retina. To correct this, a concave lens is employed. Given object distance, u = infinity (\(∞\)), and image distance, v = -80 cm. Using the lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\), we substitute the values: \(\frac{-1}{80}-\frac{1}{∞}=\frac{1}{f}\). This simplifies to \(\frac{1}{f}=-\frac{1}{80}\), resulting in a focal length f = -80 cm, or \(-0.8\) m.

The power (P) of the lens is calculated as P = \(\frac{1}{f}\). Substituting the focal length, P = \(\frac{1}{-0.8}\) = \(-1.25\ \text{D}\). Therefore, a concave lens with a power of \(-1.25\ \text{D}\) is needed to rectify the person's vision.

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