A far-sighted person has a near point of 60, | vedclass

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(C) For a far-sighted person (hypermetropia), the near point is shifted further away. The lens must form a virtual image of an object placed at the no

Vedclass · Accepted Answer

$+2.33\, D$

Vedclass · Answer

(C) For a far-sighted person (hypermetropia), the near point is shifted further away. The lens must form a virtual image of an object placed at the normal near point $(u = -25\, cm)$ at the person's actual near point $(v = -60\, cm)$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$Substituting the values: $\frac{1}{f} = \frac{1}{-60} - \frac{1}{-25}$$\frac{1}{f} = -\frac{1}{60} + \frac{1}{25} = \frac{-5 + 12}{300} = \frac{7}{300}$Power $P = \frac{100}{f(cm)} = \frac{100 	imes 7}{300} = +2.33\, D$.

$A$ far-sighted person has a near point of $60\, cm$. What power lens should be used for eyeglasses such that the person can read a book at a distance of $25\, cm$?

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