Modes of Propagation in an Optical Fibre

[TwoHats]

I'm having trouble understanding why only certain angles of propagation can transmit down an optical fibre. My lecturer produces this formula for the allowed angles:

\sin \theta = p \frac{\lambda}{2dn}

where \theta is the angle of the ray from the optical axis
\lambda the wavelength of light
d is the diameter of the fibre
n it's refractive index
and p is some integer

without any derivation saying only that the 'waves must interfere constructively'

I guess this is to do with the optical path difference being an integer number of wavelengths. However I don't understand at which point they interfere constructively nor which beams it is that are interfering.

Does this formula mean that if incident light at all angles only the angles that satisfy the above condition will emerge? Or does it mean that I must be careful to only allow these modes through else my signal will be destroyed?

 

[Andy Resnick]

I agree, it's questionable logic. The (apparent) idea is to treat light within an optical fiber as rays which reflect off the core-cladding interface. The length of the path from one side of the fiber to the other is d*sin(theta), the wavelength of light in the core is lambda/n, etc.

Unfortunately, the phrase "only certain angles of propagation can transmit down an optical fibre" is misleading, because when discussing electromagnetic propagation in waveguides it makes more sense to say "certain *modes* can transmit down an optical fibre"- the meaning is different, and modes, a wave concept, is more appropriate for spatially confined light than ray concepts.

 

[chrisbaird]

The ray is bouncing back and forth inside the fiber as it travels down: /\/\/\/\/, so it is interfering with itself. After the ray has bounced twice, it is going in the same direction and the peak of this bounced-twice ray must line up with the peak of the unbounced wave or they will destructively interfere and give you no total wave.

In reality, as Andy said, electromagnetic waves are not just rays, but are oscillations in a three-component electric and a three-component magnetic field. The more accurate picture is that of oscillating fields traveling down the waveguide, so that an optical fiber only supports modes of oscillation when the transverse part of the oscillating fields forms standing waves.

 

[myx360]

Hi, this may sound silly, but I'm struggling on a really fundamental part of this equation and its winding me up silly! I understand the fact that it's about the pathlengths and the two waves interfering all just fine but...

Why is it that d*sin(theta) is the path length? My SohCahToa says no! Isn't d the side length of the triangle opposite to the angle we're given, thus making it the denominator when we pull it over onto the same side as the sine function?

(Or to put it more clearly, I simply cannot see how d is the hypotenuse)

OP we clearly have the same lecturer, since I have been given exactly no derivation aswell.