Creative Commons License Copyright © Michael Richmond. This work is licensed under a Creative Commons License.


How to break the light up into a spectrum

All spectroscopy involves the separation of light into its individual wavelengths (or energies, or frequencies). How can one do that?

In the optical, there are two main methods. One relies upon refraction of light as it passes through a piece of glass or plastic. Remember what happens when light moves from air to glass (or vice versa)? Since the refractive index of glass is different than that of air, the light bends:

There's an equation which relates the angles of the light relative to the normal vector, before and after it enters the glass. Something about Snell ...

  Q:  What is the relationship between the angles θ1 and θ2?

Now, this isn't enough on its own to help us form a spectrum. We need to pick a medium which not only refracts light, but refracts it by a degree which depends on the wavelength of the light. In other words, we need a dispersive medium in which the index of refraction depends on the wavelength, n = n(λ) .

In common materials, the index of refraction is larger for short wavelengths. That means

  Q:  Suppose you are interested in taking the spectrum of a really,
         really faint galaxy, one for which every photon counts.
         Do you see any drawback to using a prism of glass?

Right. Refractive optics require light to pass through some material, which can cause some photons to scatter or be absorbed, and so lost.

The other main method of forming a spectrum (can) avoid this problem, because one can use a REFLECTION off a polished surface, thus preserving more photons. This second method involves diffraction gratings.

For simplicity's sake, in the diagrams below, I'll show some gratings used in a transmissive, rather than reflective, manner. It just makes things easier to see. In most astronomical applications, though, light will bounce off a grating rather than going through it.

So, a diffraction grating is a device with many grooves or little openings through which light rays can pass. Let's shine light through these openings and onto a detector, over on the right-hand side of this diagram. Consider one location on the detector, marked with an "X" in the diagram.

Now, a light ray coming from directly across from the "X" needs to travel a distance L to reach that spot on the detector. But light rays passing through an opening to one side -- a distance d or angle θ from the exact center of the grating -- must travel a slightly longer distance to reach the "X".

  Q:  How much extra distance must the second ray travel?

That extra distance is

Now, if this extra distance is equal to an integral number of wavelengths of the light, then the second ray will interfere constructively with the central ray. That means that light striking the "X" on the detector will be particularly bright. In other words, we'll see a bright spot on the detector at position "X" if

But the same is true if we look at things from a slightly different point of view. If we shine light of wavelength λ through a diffraction grating with spacing d between grooves, and the light travels a distance L before it reaches our detector, then we should see a bright spot at the very center of the detector -- straight across from the entrance, at "X"; but we should also see bright spots at off-center locations for which the angle θ satisfies this criterion.

Note that diffraction gratings will

A variety of setups for optical spectroscopy

There are a number of ways that people have arranged optics and prisms and gratings to produce spectra. Some create the spectrum of a single object; others produce spectra for hundreds of objects at once.

Spectroscopy vs. imaging: the good and the bad

What are the advantages of spectroscopy over simple imaging? Well, in a simplified way, an image tells you "how bright is this star?" But what can you learn from spectroscopy?

       Imaging                       Spectroscopy

    How bright is it?

My answers

So, if spectroscopy is so great, why don't ALL astronomers ALWAYS use spectrographs?

   Q:  Is there any advantage for a simple imaging camera
            over a spectroscopic camera attached to the same telescope?

Well, yes. If you split up the light from an object into a spectrum, then you are spreading out the light from the object across your detector. That means that the number of photons which strike each little section of your detector is much smaller ... which means that the SIGNAL-TO-NOISE must decrease.

  1. Use your browser to visit the SDSS Navigate Tool

  2. Make sure your tool is set to RA = 179.68929 and Dec = -0.454379, and the field of view is set to the "middle" setting.
  3. Roughly how many objects in this area have photometric measurements? (this requires images)
  4. Roughly how many objects in this area have spectroscopic measurements? (this requires spectra)
  5. What is the faintest object with a spectrum?
  6. What is the faintest object with a measured magnitude?
  7. To a very rough approximation, light from stars in an SDSS image was spread out over 10 pixels, and light from stars in an SDSS spectrum was spread out over 3000 pixels. What is the ratio of these areas?
  8. Convert that ratio to a magnitude difference

Basics of spectral analysis

One of the most common measurements one can make of a spectrum is the equivalent width of a line.

In the spectrum above,

  1. what is the equivalent width of the strong line centered around 486 nm?
  2. what is the equivalent width of the line centered around 489 nm?

For more information

Creative Commons License Copyright © Michael Richmond. This work is licensed under a Creative Commons License.