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.
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