Question

The far point of a myopic eye is 70 cm. Which lens should be used to correct the defect? Also find the power of the lens.

Answer 1

$<b>$
$<b><u>Given :<b><u>$


Far point of the defective eye, v = -80 cm

Object distance, u = -∞ (-infinity)

To find :

Nature and power of the corrective lens.

Solution :

1/v - 1/u = 1/f

1/f = 1/(-80) - 1/(-∞) 

1/f = - 1/80 + 0          [Since, 1/-∞ = 0]

1/f = - 1/80

f = -80 cm

Therefore, the corrective lens should be of the focal length 80 cm.

Power, P = 1 / focal length

As focal length is in centimetres, 1 m = 100 cm.

P = 100 / -80

P = -1.25 D

Therefore, the corrective lens is diverging or concave lens of power -1.5 D.