Tag: optical fibre

Questions Related to optical fibre

Optical fiber communication is based on which of the following phenomena:- 

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. Total internal reflection

  2. Scattering

  3. Reflection

  4. Interference

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Correct Option: A

If parabolic profile is used for refractive index in the core, what is the name given to such core?

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. single mode core

  2. multi mode core

  3. differential mode core

  4. curvilinear differential core

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Correct Option: D
Explanation:

Optical fibers work on the phenomenon of total internal reflection inside the fiber tube. Thin fiber tubes carry optical images to large distances,around 100 km.

The losses in the fiber is due to Raman scattering, polarization and diffraction. The loss due to diffraction can be minimized by using differential cores, that is, by having varying refractive index along the radial direction.
When these cores are parabolic, the loss is least. Such cores are called curvilinear differential cores.

What is the maximum range up to which fiber optic can be used without repeater in communication systems?

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. 4 km

  2. 10 km

  3. 100 km

  4. 500 km

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Correct Option: C
Explanation:

The maximum distance of optical link first depends on the quality of the fiber used as a medium of transmission and the insertion losses of sub-systems utilized along the link. These factors mainly limit the span the the optical repeaters required for a designed link.

limit is generally 80-100 Km and when used with amplifiers 500 km

In optical fibres, propagation of light is due to

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. diffraction

  2. total internal reflection

  3. reflection

  4. refraction

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Correct Option: B
Explanation:

Optical fibre is a device which transmits light introduced at one end to the opposite end, with little loss of the light through the sides of the fibre. It is possible with the help of total internal reflection.

An optical fibre is made of quartz filaments of refractive index 1. 70 and it has a coating of material whose refractive index is 1.45. The range of angle of incidence for one laser beam to suffer total internal reflection is

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. $0^\circ$ to $56.8^\circ$

  2. $0^\circ$ to $62.6^\circ$

  3. $0^\circ$ to $90^\circ$

  4. $0^\circ$ to $180^\circ$

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Correct Option: B
Explanation:
$i$-angle of incidence of the laser beam

$r$-angle of refraction

$i^\prime$-angle of incidence of the laser beam inside the fibre

$i _c$-critical angle

By definition of critical angle

$\displaystyle\sin{i _c}=\dfrac{1}{ _l\mu _g}=\dfrac{1}{\displaystyle\dfrac{1.70}{1.45}}=0.856$

$\implies i _c=\sin^{-1}{0.856}=58.5^\circ$

Thus if   $i^\prime>58.5^\circ\rightarrow r=90-r^\prime$

or $r<90^\circ-58.5^\circ$

$\implies r<31.5^\circ$

By snell's law, $\displaystyle\dfrac{\sin{i}}{\sin{r}}= _a\mu _g$

$\implies\sin{i}=1.70\times\sin{31.5^\circ}=1.70\times0.524=0.89$

$\implies i=\sin^{-1}{0.89}=62.6^\circ$

$\therefore$ range is $0^\circ$ to $62.6^\circ$

What should be the maximum acceptance angle at the air-core interface of an optical fibre if $\displaystyle { n } _{ 1 }$ and $\displaystyle { n } _{ 2 }$ are the refractive indices of the core and the cladding, respectively 

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. $\displaystyle { \sin }^{ -1 }\left( \frac {{ n } _{ 2 }} { { n } _{ 1 } } \right) $

  2. $\displaystyle { \sin }^{ -1 }\sqrt { { n } _{ 1 }^{ 2 }-{ n } _{ 2 }^{ 2 } }$

  3. $\displaystyle \left[ { \tan }^{ -1 }\frac { { n } _{ 2 } }{ { n } _{ 1 } } \right] $

  4. $\displaystyle \left[ { \tan }^{ -1 }\frac { { n } _{ 1 } }{ { n } _{ 2 } } \right] $

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Correct Option: B

Advantages of optical fibres over electrical wires is:

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. High band width and EM interference

  2. Low band width and EM interference

  3. High band width low transmission capacity and no EM interference

  4. High band width, high data transmission capacity and no EM interference

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Correct Option: D
Explanation:

Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it.

(A): Optical fibres are widely used to communication network.
(R) : Optical fibres are small in size, light weight, flexible and there is no scope for interference in them.

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)

  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)

  3. (A) is true but (R) is false

  4. (A) is false but (R) is true

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Correct Option: A
Explanation:

Optical fibres are widely used to communication network because they are small in size, light weight, flexible and there is no scope for interference in them. They are easy to handle due to it's small size, light weight and flexibility. They can be placed wherever we need. Further in this case, due to the no scope for interference , information cannot be loss or damage.

In optical fiber, refractive index of inner part is $1.68$ and refractive index of outer part is $1.44$. The numerical aperture of the fibre is

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

  1. 0.5653

  2. 0.6653

  3. 0.7653

  4. 0.8653

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Correct Option: D
Explanation:

$Numerical$ $aperture$ $of$ $fibre$ $=$ $\sqrt{\mu _1 ^2 - \mu _2 ^2}$ $= \sqrt{1.68^2 - 1.44^2}$

$= \sqrt{2.8224 - 2.0736}$ $= \sqrt{0.7488} = 0.8653$ 

Critical angle of glass is $\theta _1$ and that of water is $\theta _2$. The critical angle for water and glass surface would be $(\mu _g=3/2, \mu _w=4/3)$

physics refraction of light optical fibre the critical angle, total internal reflection and optical fibre total internal reflection

  1. less than $\theta _2$

  2. between $\theta _1$ and $\theta _2$

  3. greater than $\theta _2$

  4. less than $\theta _1$

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Correct Option: C
Explanation:

At critical angle. $\mu _{dense} \times sin \theta _ {dense} = \mu _{rare} \times sin 90°$


For glass - air: $(3/2) \times Sin \theta _ {1} = 1 => \theta _{1} = 41.81 ^\circ$
For water - air: $ (4/3) \times  Sin \theta _ {2} = 1 => \theta _{2} = 48.59^\circ$

For water- glass, glass is denser: $ (3/2) \times  Sin \alpha = (4/3)$ => $ \alpha = 62.72^\circ$