0
$\begingroup$

In echelle spectrographs, the presence of echelle gratings and cross-dispersers is common. What task does each one perform? Why are the cross-dispersers needed?

$\endgroup$
1
$\begingroup$

I find imagining things in three dimensions difficult and drawing them extremely difficult, an echelle spectrometer is such a thing.

Spectrographs are designed to measure the wavelengths which are present in a beam of electromagnetic radiation.

An important characteristic of a spectrograph is its resolving power which is defined as $\frac{\lambda}{\Delta \lambda}$ where $\lambda$ is the wavelength and $\Delta \lambda$ is the smallest difference from $\lambda$ which can be resolved (ie separated) by the spectrograph.

The resolving power is equal to $nN$ where $n$ is the order of the spectral line and $N$ is the number of sources (eg slits) which have been used to produce the dispersed spectrum.

Over the years the design of spectrographs has changed.
For a simple transmission grating the order of the spectrum $n$ is usually quite small, in single figures, but the total number of "slits" $N$ is comparatively high, often thousands, which makes the resolving power quite high.
Such gratings, having a large number of slits with a high precision, are expensive to produce and tend to be bulky so another way of having a large resolving power is utilised in a echelle grating.

In an echelle grating the number of sources is comparative small but the path difference undergone by the electromagnetic waves before they interfere is very large which in turn means that $N$ is small but the order of the fringes $n$ which are produced is very high.
However, there is a drawback with using such a grating on its own because there are many orders of the fringes overlapping.
That is because at a particular location where the superposition of the waves is observed there are many combinations of order $n_{\rm i}, \, n_{\rm j}$ and corresponding wavelengths $\lambda_{\rm i},\,\lambda{\rm j}$ which produce the same product $n_{\rm i}\lambda_{i} = n_{\rm j}\lambda_{j}$.
This makes interpreting an image of the spectrum very, very difficult.

enter image description here

To overcome this problem the light from a particular part of the spectrum produced by the echelle grating is split up into its component wavelengths by a cross disperser which produces a spectrum at right angles to the spectrum produced by the echelle grating.
In effect a 2D spectrum is produced whose image can be much easily be interpreted.
In other words the spectral line produced by wavelength $\lambda{\rm i}$ is now produced at a different position from that produced by wavelength $\lambda_{\rm j}$.

Here is an illustration from this paper to illustrate this 2D output from an echelle spectrograph.

enter image description here

You will note that the product of wavelength times order number on the left and right) hand side is approximately constant.

$\endgroup$
  • $\begingroup$ One of the best answers I received on this site, thank you $\endgroup$ – user3636673 Jul 27 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.