# Relating bandwidth and wavelength of a carrier

This came from a homework question, which I've solved. Just need some clarification.

I am asked to determine the bandwidth in a given amount of spectrum at a given wavelength. To find this, I know that;

$$\delta_f = \frac{c\cdot \delta_\lambda}{\lambda^2}$$

where $\delta_f$ is the bandwidth, $\delta_\lambda$ is the amount of spectrum, $\lambda$ is the given carrier wavelength, and $c$ the speed of light.

That's all well and good, but I can not find any reason for why this formula works, or where it comes from. Where does this formula come from, or what is it called, or how is it derived?

How much bandwidth is there in $0.1 \textrm{ micron}$ of spectrum at a wavelength of $1 \textrm{ micron}$?

• What do you mean by delta_lambda being "the amount of spectrum"? – Jim Clay Jul 4 '12 at 1:09
• the question reads; 'How much bandwidth is there in 0.1 micron of spectrum at a wavelength of 1 micron?" I agree it's poor wording – Michael Jul 4 '12 at 1:47
• why λl = λ - (Δλ/2)? Thank you very much in advance. – Tesla001 Sep 18 '18 at 15:51
• @Tesla001 : Welcome to SE.SP! Please do not ask questions as answers. You are better off earning enough reputation to be able to comment before doing so. – Peter K. Sep 18 '18 at 17:25

I will try to derive your formula. The formula you were given is technically not correct, but on a practical level it is. I will try to explain below.

The formula for the frequency of any wave (sound, water, light, etc.) is $f = \frac{\nu}{\lambda}$, where $\lambda$ is the wavelength and $\nu$ is the wave's velocity. This makes intuitive sense if you think about it because you will see more peaks the faster the wave travels, and you will see fewer peaks the longer the wavelength is. In your case, of course, the wave velocity is $c$.

Bandwidth is the highest frequency minus the lowest frequency.

\begin{align} B &= f_h - f_l \\ &= \frac{c}{\lambda_l} - \frac{c}{\lambda_h} \\ &= \frac{c}{\lambda - \frac{\Delta\lambda}{2}} - \frac{c}{\lambda + \frac{\Delta\lambda}{2}} \end{align} At this point to make the typing a little easier I will call $\frac{\Delta\lambda}{2}$ "$d$". \begin{align} B &= \frac{c}{\lambda - d} - \frac{c}{\lambda + d} \\ &= \frac{c(\lambda + d) - c(\lambda - d)}{(\lambda - d)(\lambda + d)} \\ &= \frac{2cd}{\lambda^2 - d^2} \\ &= \frac{c\Delta\lambda}{\lambda^2 - \frac{\Delta\lambda^2}{4}} \end{align} The formula you were given is, as a practical matter, correct, because usually $\Delta\lambda$ is much smaller than $\lambda$. In your case $\Delta\lambda$ is 1/10 of $\lambda$, which is actually much bigger than any normal situation. Usually $\Delta\lambda$ is orders of magnitude smaller than $\lambda$. Even at 1/10, though, $\frac{\Delta\lambda^2}{4}$ is $\frac{\lambda^2}{400}$, which is negligible. Thus, it is dropped.

I hope this helps.

• That helped a lot. Thank you! Makes way more sense when its derived as opposed to given in a text book. – Michael Jul 4 '12 at 5:46

The formula you were given derives from the following:

f=c/lambda.... i will call lambda =l

The derivative of f with respect to l:

df/dl = -c/l^2

Implies that df=-c*dl/l^2 The minus sign can be left out since we are only interested in the absolute value frequency of frequency variation.