The far point of a myopic eye is 250 , cm. Wh | vedclass

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(A) For a myopic eye,the far point is at a finite distance $d = 250 \, cm$. To correct this defect,the lens must form an image of an object at infinit

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$-250 \, cm$

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(A) For a myopic eye,the far point is at a finite distance $d = 250 \, cm$. To correct this defect,the lens must form an image of an object at infinity at the far point of the eye.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$u = \infty$ (object at infinity) and $v = -250 \, cm$ (image formed at the far point).Substituting these values: $\frac{1}{f} = \frac{1}{-250} - \frac{1}{\infty}$.Since $\frac{1}{\infty} = 0$,we get $\frac{1}{f} = -\frac{1}{250}$.Therefore,$f = -250 \, cm$.

$(I)$ Presbyopia	$(A)$ Sphero-cylindrical lens
$(II)$ Hypermetropia	$(B)$ Convex lens of proper power may be used close to the eye
$(III)$ Astigmatism	$(C)$ Concave lens of suitable focal length
$(IV)$ Myopia	$(D)$ Bifocal lens of suitable focal length

The far point of a myopic eye is $250 \, cm$. What is the focal length of the lens required to correct this defect?

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