Read the passage and answer the questions that follow on your understanding of the passage : Optical fibre communication is one among the new techniques have come up and are extensively used for communication purposes. The portion of the e.m. spectrum encompasses the optical region falls in the wavelength of 50 nm (Ultraviolet) to approximately 100 µm (Infrared) which includes visible light spectrum from 400 nm (Violet) to 700 nm (Red). The order of frequency corresponding to these wavelengths falls in the range 1014 to 1018 Hz. There are two types of transmission media used for optical communication i.e., atmospheric channel and guided wave channel. Out of the two, guided wave channel provide much more reliable and versatile medium of communication. The medium used for guided channels in optical range are non-metallic guides popularly known as optical fibre. The optical fibre link used for communication applications has several advantages over conventional copper cable links which makes possible to send much more data over long distances with negligible attenuation. Although some difficulties do exist in the making of optical fibre set and there are some challenges being faced yet this type of communication mechanism has made a great impact in the present scenario.
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 ?
20.2°
The acceptance angle is the inverse sine of the numerical aperture, so the whole calculation is two lines.
Step 1 — the numerical aperture.
\(NA=\sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{1.5^{2}-1.46^{2}}\)
\(=\sqrt{2.25-2.1316}=\sqrt{0.1184}=0.3441\)
Step 2 — the acceptance angle. For a fibre in air, where \(n_{0}=1\),
\(\theta_{a}=\sin^{-1}(NA)=\sin^{-1}(0.3441)=20.13^{\circ}\)
which rounds to 20.2° — option 1.
Where the formula comes from. Two conditions must hold together. A ray entering the end face refracts according to Snell's law, and once inside it must strike the core-cladding wall at more than the critical angle to be totally internally reflected:
\(\theta_{c}=\sin^{-1}\dfrac{n_{2}}{n_{1}}=\sin^{-1}\dfrac{1.46}{1.5}=76.7^{\circ}\)
Combining the two with the geometric relation between the refracted ray and the wall angle gives the numerical aperture above. Any ray entering within a cone of half-angle 20.13° about the axis is guided; any ray outside it leaks into the cladding within a few reflections.
Option 4 is a deliberate trap, and the calculation above shows exactly what it is: 76.7° is the critical angle, measured from the normal to the core-cladding boundary inside the fibre. The acceptance angle is a different quantity — measured from the fibre axis, outside, at the end face. The two are complementary in construction, not interchangeable, and confusing them is the standard error this question tests.
A check on the magnitude. The index difference here is
\(\Delta=\dfrac{1.5-1.46}{1.5}=0.027\)
which is larger than the 0.003 typical of telecom fibre, so an NA of 0.34 and a 20° acceptance cone are reasonable for a multimode fibre — wide enough to be fed by an inexpensive LED, which is what multimode step-index fibre is for.
Hence, the acceptance angle is 20.2°.
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