# How can we relate the wavelength of light to the length of a transmission line?

For the wavelength in free air we have $$\lambda = \frac{c}{f}\, [\rm{m}]$$, but how does it relate to the length of a transmission line?

My question is because of the following example from this site:

Let's say you plug your vaccuum cleaner into a wall outlet. The chord (transmission line), called $$d$$, that connects the power to the motor is $$10$$ meters long. The power is supplied at $$60$$ Hz. Should transmission line effects be taken into account?:

The wavelength at $$60$$ Hz is $$5000$$ km ($$5$$ million meters). Hence, the transmission line in this case is $$\frac{10}{5 000 000}=0.000002$$ wavelengths ($$2\cdot 10^{-6}$$ wavelengths) long. As a result, the transmission line is very short relative to a wavelength, and therefore will not have much impact on the device.

From the answer I guess we have the fraction $$\frac{d}{\lambda}=\frac{10\ \rm{m}}{5 000 000 \, \rm{m}}=0.000002 \lambda$$ But I don't follow how we can find this formula from $$\lambda = c/f$$ or why we have $$\lambda$$ at the right hand side.

TLDR: How can we relate the wavelength to the length $$d$$ of a transmission line?

• I'm not 100% sure I understood your question. You can compare any two lengths you like. For example, "this building is 85 times as tall as this barometer", or "The length of my arm is about the same as the distance travelled by light in 2.5 nanoseconds". Whether the comparison is meaningful depends on the actual physics of the situation you want to analyze. – The Photon Dec 16 '18 at 17:06

From the answer I guess we have the fraction $$\frac{d}{\lambda}=\frac{10\ \rm{m}}{5 000 000 \, \rm{m}}=0.000002 \lambda$$

That's not quite correct.

You can say

$$\frac{d}{\lambda}=0.000002$$

or

$$d = 0.000002\ \lambda.$$

In either case you'd have to calculate the value of $$\frac{10\ {\rm m}}{5,000,000\ {\rm m}}$$ to get the numerical part of the answer.

Notice, in the passage you quoted they said "the transmission line is ... 0.000002 wavelengths long." They never said the ratio of the transmission line length to the wavelength could be given with units of length.

The author of the text is making a comparison between the wavelenght $$\lambda$$ and the length of the power cord $$d$$. So it computes $$d/\lambda$$ to prove that $$d\ll\lambda$$ and the effects of wave propagation can be disregarded. Indeed you can say $$d=2\times 10^{-6}\lambda$$ with the given numbers.

I understand your question "How can we relate the wavelength of light to the length of a transmission line?". However, I am not sure I understand why you brought up that particular example from your referenced source which seems to have motivated the previous answers.

The correct relationship is $$\lambda = v/d$$ where the velocity of propagation in the transmission line is denoted as $$v$$ rather than the speed of light $$c$$. This is because when the wavelength of the signal is an appreciable factor of the length of the transmission line the line's impedance becomes a factor in effectively changing the speed of signal propagation along the line. In fact, actual RF transmission lines such as coaxial or ladder line commonly used in RF antenna systems are described as having a velocity factor. This $$v/c$$ velocity factor is used to compute $$v$$ used in the formula.

For example, if you have a typical coaxial transmission line used in antenna systems, such as RG-8X, it is published with a velocity factor $$VF=0.79$$. The formula for computing the wavelength $$\lambda$$ given frequency $$f$$ is: $$\lambda = \frac{VF\,c}{f}$$ For transmission line of RG-8X the equation is: $$\lambda = \frac{0.79\,c}{f}$$ The velocity factor that is typically published for every commercial transmission line is very important as it determines loss factors as well as effective lengths of transmission line to use in resonant circuits. If you ignore it and use the speed of light instead then your computed wavelength will be incorrect.

Lower velocity factors, such as RG58 coaxial cable where $$VF = 0.55$$ are much higher in loss with RF signals and the loss increases with increasing frequency $$f$$.

Your reference URL seems to be focused on antenna theory and how transmission lines are used in antenna systems so I think that the formula I posted using the velocity factor is an important consideration. For example, my antenna transmission line is a 450 ohm window (aka ladder) line with a published velocity factor of about 0.91 so it is a reasonably low-loss transmission line. This low loss aspect is important for my antenna, an 80-meter doublet which I use on all HF amateur radio bands (80 meter thru 10 meter bands). Since the higher frequency bands are operated with a higher SWR (standing wave ratio) due to non-resonance, the use of low loss transmission line is important.