Optical Fibres

Thread Starter


Joined Nov 23, 2006
How is the "Index of refraction of material through Prism" determined by the relation :

nm = [Sin (A+Dm)/2]
[Sin (A/2)]


Joined Feb 24, 2006
At the boundry between two materials (glass and air) a beam of light can do one of two things. It can be reflected at the boundry with the angle of reflection equal to the angle of incidence. It can also pass through the boundry and be refracted (bent) as it does so.

In total internal reflection the amount of light which escapes from the fiber is idealy zero.


Joined May 16, 2005
Are these schoolwork questions? If so, we can move this thread to the appropriate forum. If they're just musings or such, please ignore my post.


Joined Dec 13, 2006
Optical fibiers are just dielectric waveguides. Since the intrinsic impedance of the medium (the fiber) is higher than that of air, total internal reflection confines the wave (ie light) in the fiber. Outside of the fiber evenescent waves propagate, giving an exponentially decaying solution that is determined from the boundry conditions.

Ashraf Ali

Joined Nov 2, 2007
optical fibers are jus wave guides, basically they have a core (the actual guide) and a cladding surrounding it, the fiber uses Total Internal Reflection (TIR) as its principle for transmission of the intensity, TIR will occur if the refractive index of the cladding is lower than the core, the angle of incidence is greater than the critical angle (minimun angle below which TIR will not occur), and the incident intensity must be well with in the acceptance cone of the fibre. when these parameters are met TIR occurs and the wave is propogated through.