The core of an optical fiber has
a higher index of refraction than the cladding
Guidance in a fibre rests entirely on total internal reflection, and total internal reflection requires the core to be the denser medium — option 3.
\(n_{1}\gt n_{2}\qquad\text{core}\ n_{1},\ \text{cladding}\ n_{2}\)
Why the inequality must run that way. Snell's law at the core-cladding boundary gives
\(n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}\)
A critical angle exists — an angle beyond which no refracted ray can be constructed, so all the energy is reflected — only if the refracted ray bends away from the normal, that is only if \(n_{2}\lt n_{1}\). Then
\(\theta_{c}=\sin^{-1}\dfrac{n_{2}}{n_{1}}\)
and every ray striking the wall at more than \(\theta_{c}\) is reflected without loss, bouncing along the fibre indefinitely. Reverse the indices and light simply leaks into the cladding at every bounce.
| Option | Verdict |
|---|---|
| Lower than cladding | ✗ Light escapes; no guidance |
| Lower than air | ✗ Impossible — air is n = 1, the lowest of ordinary media |
| Higher than cladding | ✓ |
| Equal to air | ✗ No boundary contrast at all |
How the contrast is made. Both core and cladding are silica; the difference is created by doping — germanium or phosphorus raises the core index, fluorine or boron lowers the cladding index. The difference is deliberately tiny,
\(\Delta=\dfrac{n_{1}-n_{2}}{n_{1}}\approx0.003\)
for a typical \(n_{1}=1.48,\ n_{2}=1.46\), because a small contrast keeps the range of guided ray angles narrow and so keeps intermodal dispersion low.
Why a cladding is needed at all, given that air has a still lower index: a bare glass rod would guide light, but every speck of dust or drop of moisture touching the surface would destroy the total internal reflection at that point. The cladding keeps the guiding boundary buried inside the glass, safe from contamination and from the mechanical damage that surface scratches would cause.
Hence, the core has a higher index of refraction than the cladding.
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.
A fiber has a core radius of 6 μm, operating wavelength = 1550 nm. The V-number of the fiber is given by :
In linearly polarized modes traversing in the optical fibers the LP01 is exactly equal to :
The value of Numerical Aperture in case of optical fiber is
The core diameter of single mode fiber is in the order of
Important Questions from Optical Fiber
The material used for making optic-fibre cable in general is-
Multimode step-index fiber with a core diameter of 80 μm and a relative index difference of 1.5% is operating at a wavelength of 0.85 μm. If the core refractive index is 1.48, then the normalized frequency for the fiber is
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:
A graded indexed optical fiber has a parabolic refractive index profile (α = 2). If the fiber has a numerical aperture = 0.22 the total number of guided modes at a wavelength of 1310 nm is given by:
