Question

The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect?

• A. + 1.0D
• B. -1.0D
• C. +3.0D
• D. -3.0D

Answer 1

The correct option is C +3.0D

$Answer:-$ C option

In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence $u=-25$ cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., $v=-1m=-100$ cm.

Using lens formula, we have :-

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\frac{1}{-100}-\frac{1}{-25}=\frac{1}{f}f=\frac{100}{3}=0.33mpower=\frac{1}{f\left(inmetres\right)}=+3.0D$