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

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

  1. 4 km

  2. 10 km

  3. 100 km

  4. 500 km
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

In optical fibres, propagation of light is due to

  1. diffraction

  2. total internal reflection

  3. reflection

  4. refraction
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

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

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

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

  4. $0^\circ$ to $180^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $62.6^\circ$
Multiple choice physics option c: imaging fibre optics basics optical fibre the critical angle, total internal reflection and optical fibre

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 

  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] $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The numerical aperture (NA) is defined as sin(theta_a) = sqrt(n1^2 - n2^2). The acceptance angle is the arcsin of the numerical aperture.

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

Advantages of optical fibres over electrical wires is:

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

(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.

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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

  1. 0.5653

  2. 0.6653

  3. 0.7653

  4. 0.8653
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

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:

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

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

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

  4. $\dfrac{4Mg}{A}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stress is defined as the internal restoring force per unit area. When the wire is stretched to double its length, the instantaneous cross-sectional area decreases due to the Poisson effect, but in standard textbook problems of this type, it is assumed the force is applied to the original area or the question implies the stress at the point of doubling. However, based on standard stress-strain definitions for this specific problem type, 2Mg/A is the accepted answer.