Tag: option c: imaging

Questions Related to option c: imaging

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; $15.{ 3 }^{ 0 }$
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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. $2.525$

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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

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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

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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

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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