A fiber has a core radius of 6 μm, operating wavelength = 1550 nm. The V-number of the fiber is given by :
4.104
The V-number, or normalised frequency.
\(V=\dfrac{2\pi a}{\lambda}\,NA=\dfrac{2\pi a}{\lambda}\sqrt{n_1^{2}-n_2^{2}}\)
It is dimensionless, and it is the single parameter that decides how many modes a step-index fibre supports.
Note what the question does not give. The numerical aperture — or equivalently the core and cladding indices — is not stated, so V cannot be computed from the printed data alone. The question as reproduced is incomplete, and the missing quantity has to be recovered from the answer.
Step 1 — evaluate the geometric factor.
\(\dfrac{2\pi a}{\lambda}=\dfrac{2\pi\times6\times10^{-6}}{1550\times10^{-9}}=\dfrac{37.70}{1.55}=24.32\)
Step 2 — work backwards from each option.
| V | Required NA = V/24.32 |
|---|---|
| 2.756 | 0.113 |
| 2.956 | 0.122 |
| 4.104 | 0.1688 |
| 2.329 | 0.096 |
Step 3 — identify which NA comes from standard source data. Using the usual form \(NA=n_1\sqrt{2\Delta}\) with the textbook pair \(n_1=1.48\) and a relative index difference \(\Delta=0.65\%\):
\(NA=1.48\sqrt{2\times0.0065}=1.48\times0.1140=0.1688\)
\(V=24.32\times0.1688=4.104\)
an exact match, and the only option that corresponds to a round, standard set of fibre parameters. So option 3 is the intended answer.
What the value means. The single-mode condition is
\(V \lt 2.405\)
the first zero of the Bessel function J0. With V = 4.104 this fibre is multimode, supporting roughly
\(M\approx\dfrac{V^{2}}{2}\approx8\ \text{modes}\)
which is consistent with the unusually large 6 μm core radius — genuine single-mode fibre at 1550 nm has a core radius of about 4.1 μm, not 6.
The design lever. Since V falls as either a or NA is reduced, single-mode operation is bought with a very small core and a very small index difference — which is precisely why single-mode fibre is harder to splice and to couple light into than multimode.
Hence, the V-number is 4.104.
Similar Questions
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
Which of the following is not a usual classification of optical fibre ?
Assertion (A) : In the propagation of light along multimode graded index fibre, the rays moving toward the cladding travel longer path with greater velocity than the rays travelling shorter path near the axis of fibres. These cause less spreading as compared to spreading caused by multimode step index fibre.
Reason (R) : The velocity varies because refractive index of the multimode graded index fibre increases with radial distance from the centre (axis).
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?
Match the following lists :
List – I List – II a. \(\pi a\,NA/\lambda\) i. attenuation factor (dB/km) b. \(10\log_{10}\left(\dfrac{P_{in}}{P_{out}}\right)\) ii. Intermodal time delay c. \(\dfrac{l}{L}\dfrac{dt_g}{d\lambda}\) iii. Dispersion causing pulse spreading d. \(\dfrac{L(n_1-n_2)n_1}{n_2c}\) iv. Number of modes produced by an optical fibre Assertion (A) : Attenuation and dispersion have negative effects on the propagation of signal in the optical fibres.
Reason (R) : Optical signal degradation is caused due to structural imperfections of the fibre material.
Select your answer using the codes given below.
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