A lens of power 0.66 D is used to correct my | vedclass

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(B) For a myopic eye,the far point is at a finite distance $v$ instead of infinity. The corrective lens must form an image of an object at infinity at

Vedclass · Accepted Answer

$151.5$

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(B) For a myopic eye,the far point is at a finite distance $v$ instead of infinity. The corrective lens must form an image of an object at infinity at the person's far point.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Given power $P = -0.66 \ D$ (since it is for myopia,the lens is concave).Focal length $f = \frac{1}{P} = \frac{1}{-0.66} \ m = -\frac{100}{66} \ cm \approx -151.5 \ cm$.For an object at infinity $(u = -\infty)$,the image is formed at the far point $v$:$\frac{1}{f} = \frac{1}{v} - \frac{1}{-\infty} \implies \frac{1}{f} = \frac{1}{v} \implies v = f$.Thus,the far point is $151.5 \ cm$.

$A$ lens of power $0.66 \ D$ is used to correct myopia. What is the far point of the eye in $cm$?

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