If the least distance of distinct vision for | vedclass

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(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (f

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convex lens of focal length $87.5 \ cm$

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(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (farsightedness).To correct this,we need to use a convex lens that forms a virtual image of an object placed at $25 \ cm$ at the boy's actual near point of $35 \ cm$.Here,object distance $u = -25 \ cm$ and image distance $v = -35 \ cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.$\frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} = \frac{1}{25} - \frac{1}{35}$.$\frac{1}{f} = \frac{7 - 5}{175} = \frac{2}{175}$.$f = \frac{175}{2} = 87.5 \ cm$.Since the focal length is positive,it is a convex lens.

If the least distance of distinct vision for a boy is $35 \ cm$,then the lens to be used by the boy for correcting the defect of his eye is

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