Q. 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
A) Nearsighted, $50\ cm$
B) Farsighted, $50\ cm$
C) Nearsighted, $250\ cm$
D) Astigmatism, $50\ cm$
Explanation
(a) The defect is myopia (near sightedness)
As we know for myopic person $f = -$(defected far point)
$\Rightarrow$ Defected far point $= - f = - \frac{1}{P} = \frac{1}{( - 2)} = 0.5\ m$
$= 50\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.