Q. A given fiber has core refractive index of 1.5 and a core cladding index difference factor $(\mathrm{\Delta})$ of 0.01125. The numerical aperture is
A) 0.252
B) 0.225
C) 0.5
D) 2.25
Explanation
(b) $NA = \mu_{1}(2\mathrm{\Delta})^{1/2} = 1.5(2 \times 0.01125)^{1/2} = 0.225$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.