Optical Fiber Waveguides MCQ || Types of Optical fiber Waveguides Questions and Answers

1. Who proposed the idea of transmission of light via dielectric waveguide structure?

  1. Christian Huygens
  2. Karpon and Bockham
  3. Hondros and debye
  4. Albert Einstein

Answer: 3. Hondros and debye


The transmission of light via a dielectric waveguide structure was first proposed and  investigated at the beginning of the twentieth century. In 1910 Hondros and Debye§ conducted a theoretical study, and experimental work was reported by Schriever in 1920. However, a transparent dielectric rod, typically of silica glass with a refractive


2. Who proposed the use of clad waveguide structure?

  1. Edward Appleton
  2. Schriever
  3. Kao and Hockham
  4. James Maxwell

Answer:3. Kao and Hockham


In 1966, Kao and Hockham (Kao et. al 1966) proposed the use of clad dielectric fiber waveguide as a medium for establishing an optical communication link. The performance of a viable transmission through optical was severely restricted in the beginning by the huge loss (~1000 dB/km) provided by the optical fibers of the late 1960s.


3. Which law gives the relationship between refractive index of the dielectric?

  1. Law of reflection
  2. Law of refraction (Snell’s Law)
  3. Millman’s Law
  4. Huygen’s Law

Answer:2. Law of refraction (Snell’s Law)


Law of refraction (Snell’s Law) gives the relationship between the refractive index of the dielectric.

Snell’s law, in optics, is a relationship between the path taken by a ray of light in crossing the boundary or surface of separation between two contacting substances and the refractive index of each. This law was discovered in 1621 by the Dutch astronomer and mathematician Willebrord Snell (also called Snellius).


4. When a ray of light enters one medium from another medium, which quality will not change?

  1. Direction
  2. Frequency
  3. Speed
  4. Wavelength

Answer:2. Frequency


When a ray of light enters from one medium to another medium, its frequency and phase do not change but wavelength and velocity change.

The electric and magnetic fields have to remain continuous at the refractive index boundary. If the frequency is changed, the light at the boundary would change its phase and the fields won’t match. In order to match the field, frequency won’t change.


5. The light sources used in fibre optics communication are __________

  1. LED’s and Lasers
  2. Phototransistors
  3. Xenon lights
  4. Incandescent

Answer:1. LED’s and Lasers


Among the variety of optical sources, optical fiber communication systems almost always use semiconductor-based light sources such as light-emitting diodes (LEDs) and laser diodes because of the several advantages such sources have over the others. During the working process of optical signals they are both supposed to be switched on and of rapidly and accurately enough to transmit the signal. Also, they transmit light further with fewer errors.


6. The ratio of speed of light in air to the speed of light in another medium is called as _________

  1. Speed factor
  2. Dielectric constant
  3. Reflection index
  4. Refraction index

Answer:4. Refraction index


  • Refractive index, also called the index of refraction, is a measure of the bending of a ray of light when passing from one medium into another.
  • Refractive index is also equal to the velocity of light c of a given wavelength in empty space divided by its velocity v in a substance, or n = c/v.
  • The ratio of the speed of incident and the refracted ray in different mediums is called refractive index.


7. The ________ ray passes through the axis of the fiber core.

  1. Reflected
  2. Refracted
  3. Meridional
  4. Shew

Answer: c. Meridional


Meridional rays are rays that pass through the axis of the optical fiber. Meridional rays are used to illustrate the basic transmission properties of optical fibers. The second type is called skew rays. Skew rays are rays that travel through an optical fiber without passing through its axis. MERIDIONAL RAYS.


8. What is the numerical aperture of the fiber if the angle of acceptance is 16 degrees?

  1. 0.50
  2. 0.36
  3. 0.20
  4. 0.27

Answer:4. 0.27


The numerical aperture of fiber is related to the angle of acceptance as follows:

NA = sinθ

Where NA = numerical aperture

θ = acceptance angle.

NA = sin16

NA = 0.27


9. Light incident on fibers of angles ________ then acceptance angle do not propagate into the fiber.

  1. Less than
  2. Greater than
  3. Equal to
  4. Less than and equal to

Answer:2. Greater than


  • Light incident on fibers of angles is greater than acceptance angle does not propagate into the fiber.
  • Light is guided in the fiber by total internal reflection at the core-cladding boundary.
  • The acceptance angle of the fiber defines a cone of angles (acceptance cone) within which rays are guided in the fiber.
  • A ray of light incident within the acceptance cone of the fiber on the entrance surface of the core will impinge on the core-cladding boundary at an angle greater than the critical angle and is “trapped” inside the core, totally internally reflected at the core-cladding boundary and guided through the core.
  • Acceptance angle is the maximum angle at which light may enter into the fiber in order to be propagated. Hence the light incident on the fiber is less than the acceptance angle, the light will propagate in the fiber and will be lost by radiation.


10. _______  in optical fiber quantifying their ability to collect light and radiate outgoing light.

  1. Numerical Aperture
  2. Refractive Index
  3. Both 1 and 2
  4. None of the above

Answer:1. Numerical Aperture


The numerical aperture (N is an important parameter of optical fibers in quantifying their ability to collect light and radiate outgoing light. NA is related to the critical angle in the fiber that defines a cone of angles within which all rays are guided in the fiber by total internal Refraction

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