Q. A hypermetropic person has to use a lens of power +5 D to normalize his vision. The near point of the hypermetropic eye is
A) 1 m
B) 1.5 m
C) 0.5 m
D) 0.66 m
Explanation
(a) $f = \frac{1}{P} = \frac{1}{5}$m = 20 cm
Now, $\frac{1}{v} - \frac{1}{u} = \frac{1}{20}$
Or $\frac{1}{v} - \frac{1}{- 25} = \frac{1}{20}\ $or$\ \frac{1}{v} = \frac{1}{20} - \frac{1}{25}$
Or $\frac{1}{v} = \frac{5 - 4}{100}\ \ $or$\ \ \frac{1}{v} = \frac{1}{100}$
Or $d = 100\ $cm$\ = 1\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.