If the distance of the far point for a myopia | vedclass

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(B) For a myopia patient,the far point is at a finite distance $x$. To correct this,a concave lens is used such that the image of an object at infinit

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Double

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(B) For a myopia patient,the far point is at a finite distance $x$. To correct this,a concave lens is used such that the image of an object at infinity is formed at the far point $x$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$v = -x$ and $u = \infty$.So,$\frac{1}{f} = \frac{1}{-x} - \frac{1}{\infty} = -\frac{1}{x}$,which gives $f = -x$.If the far point distance is doubled,the new far point is $x' = 2x$.The new focal length $f'$ will be $-x' = -2x$.Since $f' = 2f$,the magnitude of the focal length becomes double.

If the distance of the far point for a myopia patient is doubled,the focal length of the lens required to cure it will become

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