Question

A person wears glasses of power -2.0 D. The defect of the eye and the far point of the person without the glasses will be: (1) Near sightedness, 50 cm (2) Far sightedness, 50 cm (3) Near sightedness, 250 cm (4) Astigmatism, 50 cm ​

Answer 1

ANSWER:

Solution

Near sightedness,  40 cm

Negative power is given, so defect of eye is nearsightedness

Also  defected  far  point  = − f = −p1​= − (−2.5) 100​ = 40  CM