Question

The near point of a hypermetropic eye is 1 m. What is the nature and power of lens required to correct the defect? (consider that the near point of the normal eye is 25 cm) ​

• A. Convex, +3 D​
• B. Concave, -5 D​
• C. Convex, +5 D​
• D. Concave, -3 D​

Answer 1

The correct option is A Convex, +3 D​

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, $u=-25cm$.
The lens used forms its virtual image at the near point of hypermetropic eye i.e., $v=-1m=-100cm$

Using lens formula, we have
$\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$

$\frac{1}{-100}$ - $\frac{1}{-25}$ = $\frac{1}{f}$

$f$ = $\frac{100}{3}$ = $0.33m$

$power=\frac{1}{f\left(inmetres\right)}$

$power=\frac{1}{0.33}$

$power=+3.0D$

Since the power of the lens is positive, the type of lens that would be uses is CONVEX LENS with a power of +3 D.