Engineering Physics Multiple Choice Questions on “Optical Fibre”.
1. What is the principle of fibre optical communication?
a) Frequency modulation
b) Population inversion
c) Total internal reflection
d) Doppler Effect
Answer: c
Clarification: In optical fibres, the light entering the fibre does not encounter any new surfaces, but repeatedly they hit the same surface. The reason for confining the light beam inside the fibres is the total internal reflection.
2. What is the other name for a maximum external incident angle?
a) Optical angle
b) Total internal reflection angle
c) Refraction angle
d) Wave guide acceptance angle
Answer: d
Clarification: Only this rays which pass within the acceptance angle will be totally reflected. Therefore, light incident on the core within the maximum external incident angle can be coupled into the fibre to propagate. This angle is called a wave guide acceptance angle.
3. A single mode fibre has low intermodal dispersion than multimode.
a) True
b) False
Answer: a
Clarification: In both single and multimode fibres the refractive indices will be in step by step. Since a single mode has less dispersion than multimode, the single mode step index fibre also has low intermodal dispersion compared to multimode step index fibre.
4. How does the refractive index vary in Graded Index fibre?
a) Tangentially
b) Radially
c) Longitudinally
d) Transversely
Answer: b
Clarification: The refractive index of the core is maximum along the fibre axis and it gradually decreases. Here the refractive index varies radially from the axis of the fibre. Hence it is called graded index fibre.
5. Which of the following has more distortion?
a) Single step-index fibre
b) Graded index fibre
c) Multimode step-index fibre
d) Glass fibre
Answer: c
Clarification: When rays travel through longer distances there will be some difference in reflected angles. Hence high angle rays arrive later than low angle rays. Therefore the signal pulses are broadened thereby results in a distorted output.
6. In which of the following there is no distortion?
a) Graded index fibre
b) Multimode step-index fibre
c) Single step-index fibre
d) Glass fibre
Answer: a
Clarification: The light travels with different speeds in different paths because of the variation in their refractive indices. At the outer edge it travels faster than near the centre But almost all the rays reach the exit end at the same time due to the helical path. Thus, there is no dispersion in the pulses and hence the output is not a distorted output.
7. Which of the following loss occurs inside the fibre?
a) Radiative loss
b) Scattering
c) Absorption
d) Attenuation
Answer: b
Clarification: Scattering is a wavelength dependent loss. Since the glass used in the fabrication of fibres, the disordered structure of glass will make some vibrations in the refractive index inside the fibre. This causes Rayleigh scattering.
8. What causes microscopic bend?
a) Uniform pressure
b) Non-uniform volume
c) Uniform volume
d) Non-uniform pressure
Answer: d
Clarification: Micro-bends losses are caused due to non-uniformities inside the fibre. This micro-bends in fibre appears due to non-uniform pressures created during the cabling of fibre.
9. When more than one mode is propagating, how is it dispersed?
a) Dispersion
b) Inter-modal dispersion
c) Material dispersion
d) Waveguide dispersion
Answer: b
Clarification: When more than one mode is propagating through a fibre, then inter modal dispersion will occur. Since many modes are propagating, they will have different wavelengths and will take different time to propagate through the fibre.
10. A fibre optic telephone transmission can handle more than thousands of voice channels.
a) True
b) False
Answer: a
Clarification: Optical fibre has larger bandwidth hence it can handle a large number of channels for communication.
11. Which of the following is known as fibre optic back bone?
a) Telecommunication
b) Cable television
c) Delay lines
d) Bus topology
Answer: d
Clarification: Each computer on the network is connected to the rest of the computers by the optical wiring scheme called bus topology, which is an application known as fibre optic back bone.
12. Calculate the numerical aperture of an optical fibre whose core and cladding are made of materials of refractive index 1.6 and 1.5 respectively.
a) 0.55677
b) 55.77
c) 0.2458
d) 0.647852
Answer: a
Clarification: Numerical aperture = (sqrt{n1^2-n2^2})
Numerical aperture = 0.55677.
13. A step-index fibre has a numerical aperture of 0.26, a core refractive index of 1.5 and a core diameter of 100micrometer. Calculate the acceptance angle.
a) 1.47°
b) 15.07°
c) 2.18°
d) 24.15°
Answer: b
Clarification: sin i = (Numerical aperture)/n
sin i = 15.07°.