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Physics: Waves, Sound, and Light - Fibre Optics and Communication Systems

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Fibre Optics is a technology that uses light pulses to transmit information through long, thin strands of glass or plastic fibres.

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The core principle behind fibre optics is Total Internal Reflection (TIR). This occurs when light travels from a medium with a higher refractive index (n1n_1) to a medium with a lower refractive index (n2n_2) at an angle of incidence greater than the critical angle (θc\theta_c).

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An optical fibre consists of three main parts: the Core (where light travels), the Cladding (which reflects light back into the core), and the Buffer Coating (protective outer layer).

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For Total Internal Reflection to occur, two conditions must be met: 1. The light must travel from a more dense medium to a less dense medium (ncore>ncladdingn_{core} > n_{cladding}). 2. The angle of incidence ii must be greater than the critical angle θc\theta_c.

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The Refractive Index (nn) is a dimensionless number that describes how fast light travels through a material, defined as n=cvn = \frac{c}{v}, where cc is the speed of light in a vacuum (3×108 m/s3 \times 10^8 \text{ m/s}).

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Advantages of Fibre Optics over traditional copper wires include higher bandwidth (more data), lower attenuation (less signal loss over distance), and immunity to electromagnetic interference (EMI).

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Communication systems using fibre optics involve a transmitter (converts electrical signals to light), the optical fibre (transmission channel), and a receiver (converts light back to electrical signals).

📐Formulae

n=cvn = \frac{c}{v}

sin⁡(θc)=n2n1\sin(\theta_c) = \frac{n_2}{n_1}

v=fλv = f \lambda

i>θc (Condition for TIR)i > \theta_c \text{ (Condition for TIR)}

💡Examples

Problem 1:

Calculate the speed of light within the core of an optical fibre if the glass used has a refractive index of n=1.5n = 1.5. Take the speed of light in a vacuum to be c=3×108 m/sc = 3 \times 10^8 \text{ m/s}.

Solution:

Using the formula n=cvn = \frac{c}{v}, we rearrange to find vv: v=cnv = \frac{c}{n} v=3×108 m/s1.5v = \frac{3 \times 10^8 \text{ m/s}}{1.5} v=2×108 m/sv = 2 \times 10^8 \text{ m/s}

Explanation:

The refractive index tells us how many times slower light travels in the medium compared to a vacuum. Since n=1.5n=1.5, light travels at 23\frac{2}{3} of its maximum speed.

Problem 2:

A light ray in an optical fibre core (n1=1.62n_1 = 1.62) hits the boundary with the cladding (n2=1.45n_2 = 1.45). Calculate the critical angle θc\theta_c.

Solution:

Using the formula sin⁡(θc)=n2n1\sin(\theta_c) = \frac{n_2}{n_1}: sin⁡(θc)=1.451.62≈0.895\sin(\theta_c) = \frac{1.45}{1.62} \approx 0.895 θc=arcsin⁡(0.895)≈63.5∘\theta_c = \arcsin(0.895) \approx 63.5^\circ

Explanation:

The critical angle is the specific angle of incidence that results in an angle of refraction of 90∘90^\circ. Any light hitting the boundary at an angle greater than 63.5∘63.5^\circ will stay inside the core.

Problem 3:

A signal is sent through a 1000 km1000 \text{ km} fibre optic cable. If the light travels at 2×108 m/s2 \times 10^8 \text{ m/s}, how long does it take for the signal to reach the destination?

Solution:

Distance d=1000 km=1×106 md = 1000 \text{ km} = 1 \times 10^6 \text{ m}. Speed v=2×108 m/sv = 2 \times 10^8 \text{ m/s}. Using t=dvt = \frac{d}{v}: t=1×1062×108t = \frac{1 \times 10^6}{2 \times 10^8} t=0.005 seconds or 5 mst = 0.005 \text{ seconds or } 5 \text{ ms}

Explanation:

Time is calculated by dividing the total distance by the speed of light in that specific medium.