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.

 

[sardar]

Thank you for your question. Let me try to explain the concept of modes of propagation in an optical fibre and how the formula provided by your lecturer relates to it.

Firstly, it is important to understand that an optical fibre is a type of waveguide that is used to transmit light signals. Light is an electromagnetic wave and it can propagate through a medium in different modes. These modes are basically different paths or patterns that the light can take while propagating through the medium.

In an optical fibre, the light propagates through the core of the fibre, which is surrounded by a cladding layer. The core has a higher refractive index compared to the cladding, which allows the light to be confined within the core and prevents it from escaping into the cladding. This is known as total internal reflection.

Now, coming to the formula provided by your lecturer, it is known as the V-number or the normalized frequency of the optical fibre. This formula relates the angle of incidence (\theta) to the other parameters such as the wavelength of light (\lambda), diameter of the fibre (d), and refractive index (n). This formula is derived from the condition of total internal reflection, where the optical path difference between the two rays of light at the core-cladding interface must be an integer multiple of the wavelength.

In simpler terms, this means that for the light to propagate through the fibre, the angle of incidence must be such that the optical path difference between the two rays is an integer multiple of the wavelength. This is why only certain angles of propagation are allowed in an optical fibre. These angles are known as the modes of propagation.

To answer your question about interference, it is the interference between these different modes of propagation that allows the light to be transmitted through the fibre. The different modes of propagation interfere constructively at certain angles, while at other angles they interfere destructively, resulting in the light being confined within the core.

So, to summarize, the formula provided by your lecturer is a way to determine the allowed modes of propagation in an optical fibre. It does not mean that you should only allow these modes, but rather these are the only modes that can propagate through the fibre. It is important to have a good understanding of these modes to effectively design and use optical fibres in various applications.

I hope this helps clarify your doubts. If you have any further questions, please feel free to ask.