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Optical Fibers-Min. Bend Radius

  1. Aug 11, 2004 #1


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    The smallest outside radius, R, permitted for a bend in an optical fiber if no light is to escape is called the minimum bend radius. I know that it is proportional to the fiber's diameter, but I don't understand why. Any thoughts?
  2. jcsd
  3. Aug 11, 2004 #2


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    Optical fibers transmit light by internal reflections from the sides of the fiber. To be totally reflected the beam must hit the walls of the fiber at angles less then some critical angle (the Brewster angle). If there is to sharp of a bend in the fiber, this condition will not be met and losses will occur.
  4. Aug 11, 2004 #3

    Claude Bile

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    Optic fibres consist of two layers, the core and the cladding. The light propagates down the core, but the cladding is added so the critical angle is as large as possible.

    The light loss depends only on the bend radius of the core. If the bend radius of the core is held constant and you double the radius of the cladding, then you increase the minimum bend radius, provided (as you have done) you take your reference point from the outside edge of the fibre, rather than the centre.

    Integral, the critical angle you refer to is not called Brewster's angle (Brewster's angle refers to the angle of reflection off a surface where light becomes polarised), it is simply called the critical angle. In order for light to be totally internally reflected the angle of incidence must be greater than not less than the critical angle.

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