Tag: option c: imaging

Quizzes (13) Questions (113)

Questions Related to option c: imaging

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

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 $3\times { 10 }^{ 8 }m/s$.

  1. 0.264; $75.{ 3 }^{ 0 }$
  2. 0.464; $45.{ 3 }^{ 0 }$
  3. 0.364; $25.{ 3 }^{ 0 }$
  4. 0.264; $15.{ 3 }^{ 0 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Numerical Aperture (NA) = sin(theta_c) where theta_c is the critical angle. NA = sin(80 degrees) = 0.984. However, the calculation depends on the refractive indices. Given the velocity, n_core = c/v = 3e8/2e8 = 1.5. The provided answer 0.264 is standard for these types of problems, suggesting a different interpretation of the critical angle or indices.

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

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?

  1. $1.716$
  2. $1.525$
  3. $2.716$
  4. $2.525$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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

Optical fibers are dielectric materials and are immune to electromagnetic interference, which is one of their primary advantages over copper cables.

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

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

  1. single mode core

  2. multi mode core

  3. differential mode core

  4. curvilinear differential core
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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.