- The use of light in digital communication.
- Understand how total internal reflection enables light to travel along a fibre.
- Perform calculations regarding TIR and critical angle.
- Understand why step-indexed fibres can be bundled together.
- Understand how waveguide dispersion is reduced by graded index fibres.
- Solve problems involving attenuation.
The use of light in digital communication
Digital devices communicate with each other with a series of 1's and 0's. This can be achieved by converting the information into an electrical signal comprising of a high potential and a low potential.
The signal is read at a constant rate so in the case above the signal would be 10100110
- What is the signal represented below?
This signal is transmitted along wires but as the wire passes from one device to another it will pass close to other electrical devices that interfere with the signal. A vacuum cleaner for example has a rotating coil that radiates a changing electric and magnetic field, this causes a changing potential in the wire which distorts the signal. An alternative is to use light.
- Read the flashing light below every second to decode the information.
The advantage of using light is that there is no interference but the problem is that light doesn't go round corners or through walls, however it can if it is totally internally reflected along a fibre.
Total internal reflection
Set up a block and laser in Algodoo as shown
- Rotate the laser until the angle of refraction = 90° this is the critical angle. (you might have to select send the block to the back in the selection menu)
- Show that the critical angle is given by
sin c = 1/n
- Surround the block with another block (send to back)
- Change the angle of the laser and and vary the refractive index of the outer block to achieve 90° refraction.
- Show that
sin c =n2/n1
Step Indexed fibre
A step index fibre has two layers like the last example above. The advantage of two layers is that you can bundle fibres together.
- Make a fibre like the one above with a core of refractive index 1.5 and cladding 1.25.
- Calculate the maximum angle of incidence for total internal reflection.
- Check your answer with the simulation.
Dispersion in step indexed fibres
There are many possible paths through fibre (modes). The image shows the longest and shortest
These rays would reach the end at different times which would mess up the signal
- Why doesn't a thin fibre (monomode) have the same problem?
Different wavelengths are refracted by different amounts.
Also much better with a monomode fibre.
Graded Index fibre
This has a cladding with progressively decreasing refractive index. Try making such a fibre in Algodoo using layers
This has many modes but remembering that light travels slowest in the most dense medium:
- Which ray travels furthest?
- Which ray travels with the highest average speed?
- Why do both rays arrive at the same time
Attenuation is the opposite of amplification, it denotes how much the signal is reduced, due to absorption, as it passes along the fibre.
attenuation = 10 log10(Power in / Power out)
Unit Decibel (dB)
- Calculate the attenuation if the power in is 100 mW and the power out is 1 mW
- If the fibre is 20 km long what is the attenuation per km?
- If the attenuation is 40 dB and the power in is 100 mW, calculate the power out.
You will have realised that the dB scale is logarithmic an attenuation of 10 x n dB gives a reduction of 1/10n