Question

The near point of a hypermetropic eye is 1 m, what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)

• A. $+3D$
• B. $-3D$
• C. $+4D$
• D. $-4D$

Answer 1

The correct option is A

$+3D$

If an object is placed at 25 cm(N'), then the image has to be formed at 100 cm(N) (because N is the far point for the hypermetropic eye), which act as the object for eye lens, which forms the final image at the retina.

$v=-1m,u=-25cm=-\frac{1}{4}m$ [sign convention according to cartesian method]

Now, using lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{-1}{1}-\frac{-4}{1}=\frac{1}{f}$
$\Rightarrow \frac{1}{f}=+3m$
Thus, power $=\frac{1}{f\text{(in metre)}}=+3D$