A person uses a lens of power +3 text{ D} to | vedclass

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(A) The power of the lens is $P = +3 	ext{ D}$.Focal length $f = \frac{1}{P} = \frac{1}{3} 	ext{ m} = \frac{100}{3} 	ext{ cm}$.For a hypermetropic

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(A) The power of the lens is $P = +3 	ext{ D}$.Focal length $f = \frac{1}{P} = \frac{1}{3} 	ext{ m} = \frac{100}{3} 	ext{ cm}$.For a hypermetropic eye,the near point is shifted beyond the normal near point $(u = -25 	ext{ cm})$. The lens forms a virtual image of an object placed at the normal near point $(u = -25 	ext{ cm})$ at the person's actual near point $(v = -x)$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{100/3} = \frac{1}{-x} - \frac{1}{-25}$$\frac{3}{100} = -\frac{1}{x} + \frac{1}{25}$$\frac{1}{x} = \frac{1}{25} - \frac{3}{100} = \frac{4-3}{100} = \frac{1}{100}$$x = 100 	ext{ cm} = 1 	ext{ m}$.

$A$ person uses a lens of power $+3 \text{ D}$ to normalize vision. The near point of the hypermetropic eye is ....... $m$.

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