Tag: physics

Questions Related to physics

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:

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

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

A cable that can support a load of 800 N is cut into two equal parts. The maximum load that can be supported by either part is 

forces on solids elastic and plastic substances forces and matter properties of matter physics

  1. 100 N

  2. 400 N

  3. 800 N

  4. 1600 N

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Correct Option: C
Explanation:

Breaking stress $= \dfrac{800 N}{A} \Rightarrow BA$

  $F = 800 N$
Breaking stress doesn't depend upon the length of the cable.

A uniform steel bar of cross-sectional area A and length L. is suspended so that it hangs vertically. The stress at the middle point of the bar is ( $\rho $ is the density of steel)

forces on solids elastic and plastic substances forces and matter properties of matter physics

  1. $\frac{L}{2A} \rho g$

  2. $\frac{L\rho g}{2} $

  3. $\frac{LA}{\rho g}$

  4. $L\rho g$

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Correct Option: C

On suspending a weight $Mg$ the length $l$ of elastic wire and area of cross section $A$ its length becomes double the initial length. The instantaneous stress action on the wire is:

forces on solids elastic and plastic substances forces and matter properties of matter physics

  1. $\dfrac{Mg}{A}$

  2. $\dfrac{Mg}{2A}$

  3. $\dfrac{2Mg}{A}$

  4. $\dfrac{4Mg}{A}$

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Correct Option: C