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?
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The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision. Object distance, u = infinity = $\infty$ Image distance, v = - 80 cm Focal length = f According to the lens formula,$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$-\frac{1}{80}-\frac{1}{\infty}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=-\frac{1}{80}$
$\text{f} = -80 \text{ cm} = -0.8\text{ cm}$
We know, Power, $\text{P}=\frac{1}{\text{f(in metres)}}$$\text{P}=\frac{1}{-0.8}=-1.25 \ \text{D}$
A concave lens of power -1.25 D is required by the person to correct his defect.
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