Fiber Optic Illumination System Research: Calculating Power & Length

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SUMMARY

This discussion focuses on the use of fiber optics for room illumination using solar power, specifically addressing the calculations needed to determine the required fiber length and quantity to achieve the equivalent power of a 60-watt lamp. Key factors include the Numerical Aperture (NA) of the fiber, which is influenced by the refractive indices of the core and cladding materials. Plastic optical fibers are recommended for their durability and ease of use. Additionally, the intensity of sunlight and the angle of incidence are crucial for estimating the light collection efficiency throughout the day.

PREREQUISITES
  • Understanding of Numerical Aperture (NA) in fiber optics
  • Knowledge of refractive indices for fiber core and cladding materials
  • Familiarity with solar intensity measurements
  • Basic principles of optical fiber design and functionality
NEXT STEPS
  • Research the formulas for calculating fiber length and quantity based on desired illumination levels
  • Learn about the properties and applications of plastic optical fibers
  • Investigate the impact of sunlight angle on fiber optic performance
  • Explore collimating optics and their role in fiber optic systems
USEFUL FOR

This discussion is beneficial for researchers, engineers, and designers interested in utilizing fiber optics for solar-powered illumination systems, as well as anyone looking to optimize light collection and distribution in architectural applications.

Aleksee
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Hello everyone,
Am having a research about the use of fiber optics in illuminating a room for example by taking solar power.
What are the formulas that are used in order to know how much length do we need and how many fibers we need in order to have for example the power of a 60 watt lamp?

I appreciate it.
Thank you.
 
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Fibers can transport quite a bit of power, the losses depend on the bends. You will need some collimating optics. Why don't you approximate the necessary sizes by the size of a window that you would put into the same room. I am sure there are good books about this.
 
As far as delivering/accepting the light:

The amount of light a fiber can collect is based on it's Numerical Aperture (NA). The math for this is based on a index of refraction of the core and a cladding that the fiber is made of. Basically the NA tells you which light will enter and stay in your fiber as a function of angle.

For transporting a large amount of light for illumination plastic optical fiber will work well enough and be much easier to work with. They are based on the same principles but much more durable, easier to work with, and cheaper. http://en.wikipedia.org/wiki/Numerical_aperture#Fiber_optics
http://en.wikipedia.org/wiki/Plastic_optical_fiber
http://www.pofeska.com/pofeskae/tece/whatspofe/whatspofe.htm

Then knowing the intensity of light from the sun in your area and the angle at which the fiber will be placed in relation to the sun you can figure out how much light to expect. This will of course vary throughout the day as the light from the sun has to go through more atmosphere and the angle of incidence to your fiber/window changes.
 
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