A person with hypermetropia has a near point | vedclass

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(D) The person has hypermetropia (farsightedness) with a near point of $60\, cm$.To read a book at $25\, cm$, the lens must form a virtual image of th

Vedclass · Accepted Answer

$+2.33\, D$

Vedclass · Answer

(D) The person has hypermetropia (farsightedness) with a near point of $60\, cm$.To read a book at $25\, cm$, the lens must form a virtual image of the object placed at $u = -25\, cm$ at the person's 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}{-60} + \frac{1}{25}$$\frac{1}{f} = \frac{-5 + 12}{300} = \frac{7}{300}\, cm^{-1}$$f = \frac{300}{7}\, cm = \frac{3}{7}\, m$Power $P = \frac{1}{f(	ext{in } m)} = \frac{1}{3/7} = \frac{7}{3} \approx +2.33\, D$.

$A$ person with hypermetropia has a near point of $60\, cm$. What power of lens should be used for the eye to read a book at $25\, cm$?

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