Question

6. An optical fiber has an acceptance angle of 10.95 degrees. What's the numerical aperture for this fiber? A. 0.16

B. 1.02

C. 0.27

D. 0.19

7. An optical fiber has a core refractive index of 1.495 and a cladding refractive index of 1.47. What's the acceptance angle for this fiber?

A. 13.11°

B. 0.272°

C. 9.79°

D. 15.80°

8. While traveling through a cable, a signal loses 18.7 percent of its original power. What's this loss expressed in dB?

A. 1.3 dB

B. 0.4 dB

C. 2 dB

D. 0.9 dB

9. An optical fiber has a bending loss of 0.5 dB per turn. You know that the fiber is bent, but you can't see how many turns it makes. You measure the bending loss to be 0.25 dB. How large is the angle used to make this bend?

A. 540°

B. 360°

C. 180°

D. 270°

10. An optical fiber has a numerical aperture of 0.22, a core diameter of 50 μm, and operates at a wavelength of 1.31 μm. How many modes can propagate in this fiber?

A. 232

B. 26

C. 695

D. 347

Answer 1

Numerical aperture:

$NA=\sqrt{n_1^2-n_2^2}$

n1- refractive index of core

n2- refractive index of cladding

$NA=sin(\alpha)$

Alpha is angle of acceptance.

With help of these formula, we can find NA.

b. alleplance angle d = 10.95 degree NA = ? NA = sind NA sin (10.95). NA = 0.189. 7 option D ti ni = 1.2195 (Core) na = 1.47 (cladding). NA = /n, ²-12² NA - /v. 495) ² - (1.4 9² INA = 0.014125 NA = 0,272 NA = sind da Sin (0.272) I d = 15.8 option D,