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 Select the correct answer from above options Electromagnetic Theory questions and answers, Electromagnetic Theory questions pdf, Electromagnetic Theory question bank, Electromagnetic Theory questions and answers pdf, mcq on Electromagnetic Theory pdf,electromagnetic theory book pdf,electromagnetic theory pdf notes,electromagnetic theory lecture notes,electromagnetic theory of light,maxwell electromagnetic theory,electromagnetic theory proposed by,electromagnetic theory engineering physics,electromagnetic theory nptel

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.