Monomode Fibres
Physics and electronics discussion, revision, exam and homework help.
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Re: Monomode FibresMonomodal fibres are very thin so light input needs to be precise otherwise it'd just refract out due to being greater than the critical angle. (Not sure if this helps much(Original post by Perpetuallity)
Quoting from my textbook "there are no zig-zag paths; light can only travel parallel to the axis, so there is no multimode dispersion. [The reason is beyond A-level]".
What exactly is the reason?
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Although a disadvantage with this would be that they can't bend as much.
On a side note - I'm also not sure if there's another method where the wire is made up of multiple layers of ever decreasing refractive indices making the critical angle larger and so light input needs to be near parallel (Not sure if there's a term for this or if I'm wrong here).
Forgot what the term is for this... -
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Re: Monomode FibresThanks, I think its called a graded fiber.(Original post by TheGrinningSkull)
Monomodal fibres are very thin so light input needs to be precise otherwise it'd just refract out due to being greater than the critical angle. (Not sure if this helps much )
Although a disadvantage with this would be that they can't bend as much.
On a side note - I'm also not sure if there's another method where the wire is made up of multiple layers of ever decreasing refractive indices making the critical angle larger and so light input needs to be near parallel (Not sure if there's a term for this or if I'm wrong here).
Forgot what the term is for this...
According to wiki, the different modes of frequency are solutions to the Helmholtz partial differential equation and I was just wondering exactly why that's the case. -
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Re: Monomode FibresThanks.(Original post by Perpetuallity)
Thanks, I think its called a graded fiber.
According to wiki, the different modes of frequency are solutions to the Helmholtz partial differential equation and I was just wondering exactly why that's the case.
Are we talking about interference now?
If we talk about having different frequencies of the signals as a solution then it should be right because interference occurs between waves of the same frequency
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