Question

The near point of a defective eye is 60 cm .Find the nature and power of the lens required

Answer 1

u = -25 cm

v = -60 cm

1 /v - 1 /u = 1 /f

1 /-60 + 1 /25 = 1 /f

f = - 60 x 25  /  25 -60

f = 300 / 7 cm

Power(P) = 1 /F in meters

P = 100 x 7 / 300

P = +2.34 D

Here , the nature of lens the person should use is concave lens.