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
\(\mu_1 = 1.42,\ \mu_2 = 1.40\)
The two defining relations for a step-index fibre.
Numerical aperture (light-gathering ability, related to the acceptance angle):
\(NA = \sqrt{\mu_1^{2}-\mu_2^{2}} = \mu_1\sqrt{2\Delta}\)
Fractional refractive-index change:
\(\Delta = \dfrac{\mu_1-\mu_2}{\mu_1}\)
The second form of NA follows from the first because \(\mu_1^{2}-\mu_2^{2}=(\mu_1+\mu_2)(\mu_1-\mu_2)\approx 2\mu_1(\mu_1-\mu_2)\) when the two indices are close, which is always the case in practical fibres.
Step 1 — find the core index.
\(\mu_1 = \dfrac{NA}{\sqrt{2\Delta}} = \dfrac{0.22}{\sqrt{2\times0.012}} = \dfrac{0.22}{\sqrt{0.024}}\)
\(\sqrt{0.024}=0.1549 \Rightarrow \mu_1 = \dfrac{0.22}{0.1549} = 1.42\)
Step 2 — find the cladding index.
\(\mu_2 = \mu_1(1-\Delta) = 1.42\times(1-0.012) = 1.42\times0.988 = 1.40\)
Step 3 — check against NA.
\(\sqrt{1.42^{2}-1.40^{2}} = \sqrt{2.0164-1.96} = \sqrt{0.0564} = 0.238 \approx 0.22\) ✓ (small rounding).
Eliminating the wrong options by inspection. Options 1 and 4 give indices of about 9.16 — physically impossible for glass, whose refractive index is around 1.44–1.48 (an index that high is not found in any optical material). Option 3 has \(\mu_1 \lt \mu_2\), i.e. the cladding denser than the core; total internal reflection would then be impossible and the fibre could not guide light at all. Only option 2 is both numerically and physically consistent.
Related idea. The acceptance angle follows from the same NA: \(\theta_{max}=\sin^{-1}(NA)=\sin^{-1}(0.22)\approx 12.7^{\circ}\), and a larger NA collects more light but increases modal dispersion, which is why long-haul fibres use a small Δ.
Hence, µ1 = 1.42 and µ2 = 1.40.
Similar Questions
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 of an optical fiber has
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:
