Q. A hypermetropic person having near point at a distance of 0.75 m puts on spectacles of power 2.5 D. The near point now is at
A) 0.75 m
B) 0.83 m
C) 0.26 cm
D) 0.26 m
Explanation
(d) $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$2.5 = \frac{1}{- 0.75} - \frac{1}{u}\ or\ \frac{1}{u} = - \frac{100}{75} - \frac{25}{10}$
Or $\ \frac{1}{u} = - \frac{4}{3} - \frac{5}{2}\ or\ \frac{1}{u} = \frac{- 8 - 15}{6} = - \frac{23}{6}$
Or $\ u = - \frac{6}{23}m = - 0.26$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.