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Fibre optics basics - class-XI

Description: fibre optics basics
Number of Questions: 20
Created by:
Tags: refraction of light physics option c: imaging ray optics and optical instruments oscillations and waves optics
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Light of different colours propagates through air :

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

  1. with velocity of air

  2. with different velocity

  3. with velocity of sound

  4. with equal velocity

Show answer
Correct Option: D
Explanation:
  • The velocity of the light does not change in air it will be equal for all the colours
  • If there may some other medium like water etc velocity will be changed
  • But it remains equal in air
  • Hence the option D is the right answer

Which of the following is used in optical fibres?

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. Diffraction

  4. Refraction

Show answer
Correct Option: A
Explanation:

In optical fibers total internal reflection property of light waves are used, so as to confine the light rays inside the fiber.

Optical fibres transmit light along its axis, by the process of :

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

  1. total internal reflection

  2. refraction

  3. interference

  4. diffraction

Show answer
Correct Option: A
Explanation:

The side-wall of optical fibre provides total internal reflection of beam of light incident on it.

In an optical fibre during transmission of light

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

  1. Energy increases.

  2. Energy decreases.

  3. No loss of propagation of energy takes place.

  4. Light partially reflects and refracts.

Show answer
Correct Option: C
Explanation:

Attenuation is a gradual loss int he intensity of any kind of flux through a medium .in telecommunication the attenuation affect the propagation of waves and signals. so during transmission of light through  an optical fiber optical and electrical attenuators cause the loss in transmission.

Which principle is the basis for the transmission of light through fiber optic cables even if the cable is bent?

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

  1. Photoelectric effect

  2. Uncertainty principle

  3. Light diffraction

  4. Light polarization

  5. Total internal reflection

Show answer
Correct Option: E
Explanation:

Optical fibres are used to transmit light from one place to another along curved path so it does not matter even if the cable is bent.Optical fibre transmission is based on phenomenon of total internal reflection .We must remember that optical fibre does not bend light, instead light follows a zig-zag path.

A cutted diamond sparkles because of its:

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

  1. high refractive index

  2. hardness

  3. emission of light by the diamond

  4. absorption of light by the diamond

Show answer
Correct Option: A
Explanation:

A cutted diamond sparkles because of its high refractive index. Because of high refractive index, the light suffers total internal reflections many times. that is why it sparkles.

The velocity of light in the core of a step index fibre is $2\times { 10 }^{ 8 }m/s$ and the critical angle at the core-cladding interfere is ${ 80 }^{ 0 }$. Find the numerical aperture and the acceptance angle for the fibre in air. The velocity of light in vacuum is $3\times { 10 }^{ 8 }m/s$.

physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre
  1. 0.264; $75.{ 3 }^{ 0 }$
  2. 0.464; $45.{ 3 }^{ 0 }$
  3. 0.364; $25.{ 3 }^{ 0 }$
  4. 0.264; $15.{ 3 }^{ 0 }$
Show answer
Correct Option: D

The optical path of a monochromatic light is the same if it goes through $2.00$ cm of glass or x cm of ruby. If the refractive index of glass is $1.510$ and that of ruby is $1.760$ find the value of x is _______ cm?

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

  1. $1.716$

  2. $1.525$

  3. $2.716$

  4. $2.525$

Show answer
Correct Option: A
Explanation:

We know optical path =Refractive index $\mu \times$ length $x$=constant

Therefore ${ \mu  } _{ glass }{ x } _{ glass }={ \mu  } _{ ruby }{ x } _{ ruby }\ { x } _{ ruby }=\dfrac { { \mu  } _{ glass }{ x } _{ glass } }{ { \mu  } _{ ruby } } =\frac { 1.51\times 2 }{ 1.76 } =1.716$

Consider telecommunication through optical fibres. Which of the following statements is NOT true? 

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

  1. Optical fobres can be of graded refractive index

  2. Optical fibres are subjected to electromagnetic interference from outside

  3. Optical fibres have extremely low transmission loss

  4. Optical fibres may have homogeneous core with a suitable cladding

Show answer
Correct Option: B

Optical fibre, which is used in communication, works on which principle?

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

  1. Dispersion

  2. Total Internal reflection

  3. Reflection

  4. Refraction

Show answer
Correct Option: B

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

Show answer
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

Show answer
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

Show answer
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

Show answer
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 $180^\circ$

Show answer
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] $

Show answer
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

Show answer
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

Show answer
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

Show answer
Correct Option: D
Explanation:

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

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

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