The core diameter of single mode fiber is in the order of
10 µm
A single-mode core is about 8 to 10 µm across — option 2 — and the figure is not arbitrary; it is fixed by the condition for single-mode operation.
The number of guided modes is governed by the normalised frequency,
\(V=\dfrac{2\pi a}{\lambda}\sqrt{n_{1}^{2}-n_{2}^{2}}=\dfrac{2\pi a}{\lambda}\,NA\)
and only one mode propagates when
\(V\lt2.405\)
the first zero of the Bessel function \(J_{0}\). Put the numbers in for the 1.55 µm window with NA = 0.12:
\(a\lt\dfrac{2.405\times1.55}{2\pi\times0.12}=4.9\ \mu\text{m}\)
a radius of about 5 µm, so a diameter near 10 µm. The answer follows from the operating wavelength and nothing else — which is why the core cannot be shrunk indefinitely.
| Option | Size | Verdict |
|---|---|---|
| 100 µm | Multimode / plastic fibre | ✗ V ≈ 24, many modes |
| 10 µm | Single mode | ✓ |
| 1 Å | 0.1 nm — an atom | ✗ Absurdly sub-wavelength |
| 1 nm | A few atoms | ✗ Same objection |
Options 3 and 4 fail on physical grounds, not just numerically. A waveguide much smaller than the wavelength — 1550 nm here — guides nothing; the field is not confined but radiates away. An Ångström is roughly one atomic diameter, so such a "core" could not be built even in principle.
What the small core buys. Only one path exists, so intermodal dispersion vanishes entirely and the bandwidth-distance product rises from a few tens of MHz·km for step-index multimode fibre to tens of THz·km. What it costs is handling: a 9 µm core demands a laser rather than an LED, fusion splicing with sub-micron alignment, and connectors held to comparable tolerance — the reason multimode fibre with its 50 or 62.5 µm core survives for short in-building links.
Hence, the core diameter is of the order of 10 µm.
The core of an optical fiber has
Consider the following statements :
Losses in optical fibers are caused by
1. Impurities in the fibre material
2. Microbending
3. Splicing
4. Step index profile
Of these statements :
Assertion (A) : Optical fibers have broader bandwidth compared to conventional copper cables.
Reason (R) : Low power LASER beams are considered to be very powerful as compared to high power ordinary light beams.
The following is true for the multimode graded index fiber :
1. The refractive index varies as a function of radial distance from the centre.
2. The refractive index undergoes sudden change at the cladding boundary.
3. It provides better bandwidth and the data rate than the multimode step index.
4. It provides the better bandwidth and data rate than single mode step index.
A multimode step-index fibre has glass core (n1 = 1.5) and fused quartz cladding (n2 = 1.46), which one of the following is the value of acceptance angle ?
Following is not the usual classification of an optical fibre :
Which of the following are the cases of signal attenuation ?
1. Splicing
2. Intermodal Delay
3. Scattering
4. Chromatic Dispersion
(A) Multimode fibre is less lossy than single mode
(B) The bandwidth of step index fibre is 50 MHz
(C) The graded index fibre has theoretically infinite bandwidth
(D) The step index fibre has numerical aperture of 0.2 to 0.5
(E) The graded index fibre has numerical aperture of 0.46 to 0.99
Choose the most appropriate answer from the options given below :
If numerical aperture and fractional refractive index of an optical fibre are 0.22 and 0.012, respectively. The refractive index of core (µ1) and cladding (µ2) will be
An optical fibre has numerical aperture (NA) of 0.3 and refractive index $\eta_2$ of cladding material is 1.6. What is the refractive index of core material?
What is the relation between the refractive index of core n1 and cladding n2?
Graded index fiber is used to
In a multimode fiber (step index), number of modes passing at an operating wavelength of 1300 nm are 1000, the refractive index of the core is 1.50 and that of the cladding is 1.48. The value of core diameter is:
In optical fibers, the Rayleigh scattering is proportional to:
Fibre optic power meters have input for attaching fiber optic connector and detector: