Cylindrical Fiber MCQ || Optical fiber Questions and Answers

Optical Fiber, Electronics MCQ / March 15, 2022

1. Multimode step-index fibers allow the propagation of a _______ number of guided modes along the channel.

  1. Finite
  2. Infinite
  3. Equal
  4. All of the above

Answer:1. Finite

Explanation: 

  • Multimode step-index fibers allow the propagation of a finite number of guided modes along the channel.
  • The number of guided modes is dependent upon the physical parameters (i.e., relative refractive index difference, core radius) of the fiber and the wavelengths of the transmitted light, which are included in the normalized frequency V for the fiber.

The total number of guided modes or mode volume M, for a step-index fiber, is related to the V value for the fiber by the approximate expression:

M= V2/2

Where V = Normalize Frequency

It allows an estimate of the number of guided modes propagating in a particular multimode step-index fiber.

 

2. A multimode step-index fiber has a normalized frequency of 72. Estimate the number of guided modes.

  1. 2846
  2. 2592
  3. 2432
  4. 2136

Answer:2. 2592

Explanation: 

  • Multimode step-index fibers allow the propagation of a finite number of guided modes along the channel.
  • The number of guided modes is dependent upon the physical parameters (i.e., relative refractive index difference, core radius) of the fiber and the wavelengths of the transmitted light, which are included in the normalized frequency V for the fiber.

The total number of guided modes or mode volume M, for a step-index fiber, is related to the V value for the fiber by the approximate expression:

M= V2/2

Here M denotes the number of modes and V denotes normalized frequency.

M = (72)2/2 = 5184/2

M = 2592

 

3. Skew rays follow a ______

  1. Hyperbolic path along the axis
  2. Parabolic path along the axis
  3. Helical path
  4. Path where rays change angles at the core-cladding interface

Answer: 3. Helical path

Explanation: 

Skew rays do not pass through the fiber axis. They follow the helical path in the fiber core.

Skew-ray

  • The skew ray is reflected at the edges of the fiber core with an angle of 2γ.
  • The number of reflections of the skew rays depends upon the point of incidence and angle of incidence in the core.
  • If the input light is nonuniform, then skew rays follow smooth reflections, and more uniform output is obtained.

 

4. A graded-index fiber has a core with a parabolic refractive index profile of diameter of 30μm, NA=0.2, λ=1μm. Estimate the normalized frequency.

  1. 19.32
  2. 18.84
  3. 16.28
  4. 17.12

Answer: 2. 18.84

Explanation: 

Normalized frequency is a unit of measurement of frequency equivalent to cycles/samples.

Normalized frequency for a graded-index fiber is given by

V = 2πa(N/λ.

where

a is the core radius,= 30/2 = 15μm

λ is the wavelength in vacuum = 1 μm

V is the normalized frequency

NA = numerical aperture = 0.2

Substituting and calculating the values

V = 2 × π × 15 × 0.2/1

V = 18.84

 

5. A step-index fiber has a core refractive index of 1.46 and a radius of 4.5μm. Find the cutoff wavelength to exhibit single-mode operation. Use relative index difference as 0.25%.

  1. 1.326μm
  2. 0.124μm
  3. 1.214μm
  4. 0.123μm

Answer: 3. 1.214μm

Explanation: 

A step-index fiber is fiber type cylindrical waveguide core with an inner core has a uniform refractive index of n1 and the core is surrounded by an outer cladding with a uniform refractive index of n2.

Calculation

cut off wavelength is given by

$\lambda _{c}=\frac{2\Pi an_{1}\sqrt{2\Delta }}{V_{c}}$

For cut-off wavelength, Vc = 2.405

$\lambda _{c}=\frac{2\Pi \times 4.5\times 1.46\sqrt{0.005 }}{2.405}$

λc = 1.214 μm

 

6. A step-index fiber has a core refractive index of 1.46 and a radius of 4.5μm. Find the Numerical Aperature to exhibit single-mode operation. Use relative index difference as 0.25%.

  1. 0.707
  2. 1.085
  3. 2.4
  4. 707

Answer: 1. 0.707

Explanation: 

Given

core radius,= 4.5 μm

Refractive index difference Δ0.25% = 0.0025

Core refractive index n1 = 1.46

Numerical Aperature = n1(2Δ)1/2

= 1.46(2× 0.0025)1/2

NA =  0.707

 

7. A single-mode step-index fiber allows propagation of only one transverse ________

  1. Magnetic wave
  2. Electrical Wave
  3. Electromagnetic wave
  4. None of the above

Answer: 3. Electromagnetic wave

Explanation: 

  • Single-mode step-index fiber allows the propagation of only one transverse electromagnetic mode.
  • The core diameter is of the order of 2 to 10 µm.
  • As single-mode fibers support only one mode, these types of fibers do not exhibit Intermodal dispersion, whereas with multimode step-index fibers considerable dispersion may take place due to different group velocities of the propagating modes.

 

8.  The core diameter of the single-mode step-index fiber is

  1. 10 to 20 µm
  2. 2 to 10 µm
  3. 20 to 40 µm
  4. 50 to 100 µm

Answer: 2. 2 to 10 µm

Explanation: 

  • Single-mode step-index fiber allows the propagation of only one transverse electromagnetic mode.
  • The core diameter is of the order of 2 to 10 µm.
  • As single-mode fibers support only one mode, these types of fibers do not exhibit Intermodal dispersion, whereas with multimode step-index fibers considerable dispersion may take place due to different group velocities of the propagating modes.

 

9. Which of the following statements is true for intermodal dispersion.

  1. Low in single-mode and considerable in multimode fiber
  2. Low in both single-mode and multimode fiber
  3. High in both single-mode and multimode fiber
  4. High in single-mode and low in multimode fiber

Answer: 1. Low in single-mode and considerable in multimode fiber

Explanation: 

  • Single-mode step-index fiber allows the propagation of only one transverse electromagnetic mode.
  • The core diameter is of the order of 2 to 10 µm.
  • As single-mode fibers support only one mode, these types of fibers do not exhibit Intermodal dispersion, whereas with multimode step-index fibers considerable dispersion may take place due to different group velocities of the propagating modes.

 

10. For lower bandwidth applications _______

  1. Single-mode fiber is advantageous
  2. Photonic crystal fibers are advantageous
  3. Coaxial cables are advantageous
  4. Multimode fiber is advantageous

Answer: 4. Multimode fiber is advantageous

Explanation:

The single-mode step-index fiber has the distinct advantage of low intermodal dispersion (broadening of transmitted light pulses), as only one mode is transmitted

With multimode step-index fiber, a considerable dispersion may occur due to the different group velocities of the propagating modes. This, in turn, restricts the maximum bandwidth attainable with multimode step-index fibers, especially when compared with single-mode fibers.

In multimode fibers, intermodal dispersion occurs. The group velocities often differ which gradually restricts maximum bandwidth attainability in multimode fibers.

However, for lower bandwidth applications, multimode fibers have several advantages over single-mode fibers.

  • The use of spatially incoherent optical sources (e.g., most light-emitting diodes) cannot be efficiently coupled to single-mode fibers.
  • Larger numerical apertures, as well as core diameters, facilitate easier coupling to optical sources.
  • Lower tolerance requirements on fiber connectors.

 

11. Most of the optical power is carried out in the ______  region than in ______.

  1. Core, Cladding
  2. Cladding, Core
  3. Core, Cable jacket
  4. None of the above

Answer: 1. Core, Cladding

Explanation: 

Most of the optical power is carried out in core region than in cladding.

Cladding in optical fibers is one or more layers of materials of lower refractive index, in intimate contact with a core material of higher refractive index. The cladding causes light to be confined to the core of the fiber by total internal reflection at the boundary between the two.

The core of a conventional optical fiber is the part of the fiber that guides the light. It is a cylinder of glass or plastic that runs along the fiber’s length. The core is surrounded by a medium with a lower index of refraction, typically a cladding of different glass, or plastic.

In an ideal multimode fiber, there is no mode of coupling. The optical power launched into a particular mode remains in that mode itself. The majority of these modes are mostly confined to fiber core only.

 

12. Meridional rays in graded-index fibers follow ______

  1. Straight path along the axis
  2. Curved path along the axis
  3. Path where rays change angles at core-cladding interface
  4. Helical path

Answer: 3. Curved path along the axis

Explanation: 

Meridional rays pass through the axis of the fiber core. Meridional rays shown appear to follow curved paths through the fiber core. This may be described using the concepts of geometrical optics. The gradual decrease in refractive index from the center of the core creates a large number of refractions of the rays as they are effectively incident on many high to low index interfaces.

Meridional rays

This ray indicates the transmission of light in a perfect optical fiber.

Meridional rays can be of two types :

(i) Bound rays

(ii) Unbound rays.

The bound rays follow total internal reflection in the fiber core and do not escape out of the fiber core.

Unbound rays are refracted out of the fiber core and are absorbed in the cladding.

 

13. What is the unit of normalized frequency?

  1. Hertz
  2. Meter/sec
  3. Coulombs
  4. It is a dimensionless quantity

Answer: 4. It is a dimensionless quantity

Explanation: 

The normalized frequency of optical fiber is the frequency that exists at the cut-off condition. There is no propagation and attenuation above the cut-off. It is directly proportional to the numerical aperture which is a dimensionless quantity; hence itself is a dimensionless quantity.

Normalized frequency for a graded-index fiber is given by

V = 2πa(N/λ.

where

a is the core radius

λ is the wavelength in vacuum

V is the normalized frequency

NA = numerical aperture

 

14. Multimode graded-index fibers accept ______ than multimode step-index fibers with the same core.

  1. Less light
  2. More Light
  3. No light
  4. None of the above

Answer: 1. Less light

Explanation: 

Multimode graded-index fibers accept less light than multimode step-index fibers with the same core A. However, graded-index fibers usually outperform the step-index fibers. The core’s parabolic refractive index profile causes multimode graded-index fibers to have less modal dispersion.

 

15. The advantages of skew rays over meridional rays are

  1. Large Acceptance angle
  2. High light gathering capability
  3. Both 1 and 2
  4. None of the above

Answer: 3. Both 1 and 2

Explanation: 

Advantages of skew Rays

1. Skew rays have larger acceptance angles than meridional rays.

2. Light gathering capability of skew rays is higher than meridional rays.

Disadvantage

1. Skew rays propagate in the annular region near the outer surface of the core. Hence full fiber core is not utilized.

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