Question

The near point and far point of a person are $40\, cm$ and $250 \,cm$ respectively. Determine the power of the lens he/she should use while reading a book kept at distance $25 \,cm$ from the eye.

• A. 2.5 D
• B. 5.0 D
• C. 1.5 D
• D. 3.5 D

Answer 1

Answer: (C)

1.5 D

Answer 2

If the object is placed at a distance $25 \,cm$ from the corrected lens, it should produce the virtual image at $40\, cm$. Thus, $u=-25\, cm$ and $v=-40\, cm$ The lens formula is $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\Rightarrow \frac{1}{f} =-\frac{1}{40}+\frac{1}{25}$ $\Rightarrow f =\frac{200}{3} cm =\frac{2}{3} m$ $\therefore P =\frac{1}{f}=\frac{1}{\frac{2}{3}}=\frac{3}{2}$ $=+1.5 \,D$