Question

A hypermetropia eye is corrected for near vision by using a lens of power 8/3 d. Calculate how far is the near point of defective eye. Also draw the ray of diagram to show the correction made with a suitable lens for hypermetropic eye.

Answer 1

u = -25

v = -75

therefore,

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

$\frac{1}{f} = \frac{1}{ - 75} - \frac{1}{ - 25}$

$f = \frac{75}{2} cm$

$p = \frac{100}{f}$

$p = \frac{100}{ \frac{75}{2} }$

$= + \frac{8}{3}$

D= +8/3

Answer 2

Answer:

Explanation:

P= 1/f

8/3=1/f

8f=3

f=3/8

1/f = 1/v-1/u

8/3= 1/v-1/-25

8/3 + 1/-25 =1/v

-8+6/-150=1/v

-2/-150=1/v

1/v=1/75

v=75

hence,v=-75