The near point of a hypermetropic person is 7 | vedclass

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Question

(A) For a hypermetropic person,the near point is $d' = 75\, cm$. The person wants to read a book at the normal near point $d = 25\, cm$.Using the lens

Vedclass · Accepted Answer

$+ 2.67\, D$

Vedclass · Answer

(A) For a hypermetropic person,the near point is $d' = 75\, cm$. The person wants to read a book at the normal near point $d = 25\, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$v = -75\, cm$ (the image must be formed at the person's actual near point) and $u = -25\, cm$ (the object distance).$\frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1 + 3}{75} = \frac{2}{75}\, cm^{-1}$.The power $P$ in Diopters is given by $P = \frac{100}{f(cm)}$.$P = 100 	imes \frac{2}{75} = \frac{200}{75} = +2.67\, D$.

The near point of a hypermetropic person is $75\, cm$ from the eye. What is the power of the lens required to enable the person to read clearly a book held at $25\, cm$ from the eye?

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