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?

Question

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? (A) +2 D (B) +2.5 D (C) +2.67 D (D) +3.33 D

Tardigrade Admin · Accepted Answer

u=-25 cm , v=-75 cm (1/f)=(1/v)-(1/u)=(1/(-75))-(1/(-25)) =(1/-75)+(3/75)=(2/75 cm ) P=(1/f( m ))=(2/0.75 m )=+2.67 D

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

+2 D

+2.5 D

+2.67 D

+3.33 D

$u = - 25 c m, v = - 75 c m$
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{( - 75 )} - \frac{1}{( - 25 )}$
$= \frac{1}{- 75} + \frac{3}{75} = \frac{2}{75 c m}$
$P = \frac{1}{f ( m )} = \frac{2}{0.75 m} = + 2.67 D$

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