A person suffering from hypermetropia cannot | vedclass

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(D) The person has a near point of $u = -120 \, cm$ due to hypermetropia.To read at a distance of $v = -40 \, cm$,the lens must form a virtual image o

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$-60$

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(D) The person has a near point of $u = -120 \, cm$ due to hypermetropia.To read at a distance of $v = -40 \, cm$,the lens must form a virtual image of an object placed at $40 \, cm$ at the person's actual near point of $120 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-40} - \frac{1}{-120}$.$\frac{1}{f} = -\frac{1}{40} + \frac{1}{120} = \frac{-3 + 1}{120} = \frac{-2}{120} = -\frac{1}{60}$.Therefore,$f = -60 \, cm$.

$(I)$ 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

$A$ person suffering from hypermetropia cannot focus on objects closer than $120 \, cm$. To read at a distance of $40 \, cm$,what should be the focal length of the lens required?

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