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Angles of diffraction: the grating equation

Figure 4: A collimated beam incident from the left on a reflection grating and the outgoing diffracted beams (red and blue). The incident and diffracted angles $\alpha $ and $\beta $ are governed by the grating equation and depend on wavelength and the lines/mm of the grating.

A transmission grating deviates light of wavelength $\lambda$ by an angle $\alpha $,

\begin{displaymath}{\rm sin} \alpha = k_{order} \lambda / l_{grating}, \end{displaymath}

where $l_{grating}$ = interline spacing, and $k_{order}$ is an integer $\geq 1$. Equivalently,

\begin{displaymath}{\rm sin} \alpha = k_{order} \lambda M_{grating}. \end{displaymath}

Let's work in first order where $k_{order} = 1$. For a reflection grating the grating equation is

\begin{displaymath}{\rm sin} \alpha + {\rm sin} \beta = k_{order} \lambda M_{grating} \end{displaymath}

where $\alpha $ and $\beta $ are the angles of incident and diffracted rays with respect to the grating normal, shown in Figure 4. The diffracted beams are shown as red and blue. Light from a single point source produces one incident collimated beam. The diffracted beam of a single point source at a single wavelength (e.g. the blue lines) is still collimated and will be imaged at a single point on the detector, while the red lines will be imaged at a different point. The fact that wavelengths are separated in angle, but each single wavelength stays collimated, is why we want to have the disperser in the collimated beam.

The zeropoint of the diffracted angle $\beta $ depends on the incident angle and grating normal, but the change in $\beta $ with $\lambda$ governs the resolution of the spectrograph.

We're not directly concerned with the zeropoint of $\beta $, assuming we have tilted the grating so as to get light into the camera, but with the change in $\beta $ per wavelength and the resulting wavelength scale per pixel. Consider how the camera translates a deviation in angle of the diffracted beam to a distance on the detector, $r_{ccd}$. For a small deviation in angle, $d\beta $, the image moves

\begin{displaymath}dr_{ccd} = f_{cam}  d\beta. \end{displaymath}

By differentiating the grating equation, for a fixed input angle $\alpha $,

\begin{displaymath}{\rm cos} \beta d\beta = M_{grating} d\lambda.\end{displaymath}

For typical spectrograph layouts (other than echelles), $\beta $ is not large and ${\rm cos} \beta$ is slightly $<1$.

This gives the wavelength/physical scale at the CCD:

\begin{displaymath}\frac{d\lambda}{dr_{ccd}} = \frac{{\rm cos} \beta}{f_{cam} M_{grating}}. \end{displaymath}

next up previous
Next: Spectrograph resolution Up: Spectroscopy Previous: Spectroscopy
Benjamin Weiner 2008-10-03