Question

The near point of a hypermetropic eye is 0.5 m .find the power of the lens required to correct this defect given that the near point of the normal eye 0.25m. (PLEASE REPLY THIS QUESTION IT IS URGENT )​

Answer 1

Near point of hypermetropic eye = 0.5m = 0.5 x 100 = 50cm

Near point of normal eye = 0.25m =0.25 x 100 = 25cm

So,

v = -50cm

u = 25cm

So by lens formula:-

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

$\frac{1}{f} = \frac{1}{-50} + \frac{1}{25}\\\frac{1}{f} = \frac{-1}{50} + \frac{2}{50}\\\frac{1}{f} = \frac{1}{50}$

So f = 50cm = 0.5m

So P = $\frac{1}{f}$

So P = $\frac{1}{0.5}$

So P = +2D