This set of Optical Communications Questions and Answers for Freshers focuses on “Material Absorption & Fiber Bend Losses In Silicon Glass Fibers”.
1. Which of the following statements best explain the concept of material absorption?
a) A loss mechanism related to the material composition and fabrication of fiber.
b) A transmission loss for optical fibers.
c) Results in attenuation of transmitted light.
d) Causes of transfer of optical power
Explanation: Material absorption is a loss mechanism which results in dissipation of transmitted optical power as heat in a waveguide. It can be caused by impurities or interaction with other components of core.
2. How many mechanisms are there which causes absorption?
Explanation: Absorption is a loss mechanism. It may be intrinsic, extrinsic and also caused by atomic defects.
3. Absorption losses due to atomic defects mainly include-
b) Missing molecules, oxygen defects in glass
c) Impurities in fiber material
d) Interaction with other components of core
Explanation: Atomic defects are imperfections in the atomic structure of fiber material. Atomic structure includes nucleus, molecules, protons etc. Atomic defects thus contribute towards loss of molecules, oxygen etc.
4. The effects of intrinsic absorption can be minimized by-
c) Suitable choice of core and cladding components
Explanation: Intrinsic absorption is caused by interaction of light with one or more components of the glass i.e. core. Thus, if the compositions of core and cladding are chosen suitably, this effect can be minimized.
5. Which of the following is not a metallic impurity found in glass in extrinsic absorption?
Explanation: In the optical fibers, prepared by melting techniques, extrinsic absorption can be observed. It is caused from transition metal element impurities. In all these options, Si is a constituent of glass and it cannot be considered as an impurity to glass itself.
6. Optical fibers suffer radiation losses at bends or curves on their paths. State true or false
Explanation: Optical fibers suffer radiation losses due to the energy in the bend or curves exceeding the velocity of light in cladding. Hence, guiding mechanism is inhibited, which in turn causes light energy to be radiated from the fiber.
7. In the given equation, state what αr suggests;
a) Radius of curvature
b) Refractive index difference
c) Radiation attenuation coefficients
d) Constant of proportionality
Explanation: Above equation represents the fiber loss. This loss is seen at bends and curves as the fibers suffer radiation losses at curves. These radiation losses are represented by a radiation attenuation coefficient (αr).
8. A multimode fiber has refractive indices n1= 1.15, n2=1.11 and an operating wavelength of 0.7μm. Find the radius of curvature?
Explanation: The radius of curvature of the fiber bend of a multimode fiber is given by
Where, Rc = radius of curvature
n1, n2= refractive indices
9. A single mode fiber has refractive indices n1=1.50, n2= 2.23, core diameter of 8μm, wavelength=1.5μm cutoff wavelength= 1.214μm. Find the radius of curvature?
a) 12 mm
b) 20 mm
c) 34 mm
d) 36 mm
Explanation: The radius of curvature of the fiber bend of a single mode fiber is given by-
Where R= radius of curvature,
n1, n2= refractive indices,
λc= cutoff wavelength,
λ= operating wavelength.
10. How the potential macro bending losses can be reduced in case of multimode fiber?
a) By designing fibers with large relative refractive index differences
b) By maintaining direction of propagation
c) By reducing the bend
a) By operating at larger wavelengths
Explanation: In the case of multimode fibers, radius of curvature is directly proportional to core refractive index and operating wavelength. In order to reduce the macro bending losses, the operative wavelength must be small and fibers must have large relative refractive index difference. Losses are inversely proportional to refractive index differences.
11. Sharp bends or micro bends causes significant losses in fiber. State true or false
Explanation: Sharp bends usually have a radius of curvature almost near to the critical radius. The fibers with the radius near to the critical radius cause significant losses and hence they are avoided.
Sanfoundry Global Education & Learning Series – Optical Communications.
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