Question

The numerical aperture of a coaxial cable with core and cladding indices given by 2.33 and 1.4 respectively is

(a) 3.73

(b) 0.83

(c) 3.46

(d) 1.86

I have been asked this question during an interview.

Asked question is from Refractive Index and Numerical Aperture in section EM Wave Propagation of Electromagnetic Theory

Answer 1

Correct option is (d) 1.86

For explanation: The numerical aperture is given by NA = √(n1^2 – n2^2), where n1 and n2 are the refractive indices of core and cladding respectively. On substituting for n1 = 2.33 and n2 = 1.4, we get NA = √(2.33^2-1.4^2) = 1.86.