We can photograph spectra using the Phasmatron spectroscope in two ways:

- Direct method
- Inverse eyepiece projection method

With the direct method, we remove the viewing telescope, and place a camera with a lens of focal length f. With the eyepiece projection method, we place the camera in front of the eyepiece of the viewing telescope, and project the image into the camera, without removing the camera lens. Since the camera lens and the eyepiece do not fit exactly, a dark piece of cloth (black) should be placed around the junction point, so that no external light is collected by the camera. Here is how we would calculate for example the image dispersion with the two different methods. On the neighborhood of the blue Mercury line (4358.35A), we will do the following calculation:

- Direct method photography, using an f=50mm lens:
dn/dë=(1.76197-1.74805)/|4358.35-4799.9107|=3.1524545*10

^{-5}/A.dE/dn=2/cos{arcsin{n

_{4358.35}/2}}=4.22704rad.dE/dë=4.22704rad*3.1524545*10

^{-5}/A=1.3325551*10^{-4}rad/A.ds/dë=50*1.3325551*10

^{-4}rad/A=6.66277*10^{-3}mm/A.The last value is equivalent to dë/ds=150A/mm. This means that our scale in the area of the Mercury blue line must have roughly 150A for every mm, on film. Given the fact that most times the film is enlarged, we should adjust by this value. If our film is enlarged 5 times, then we get 30A/mm.

- Inverse eyepiece projection method using the Phasmatron spectroscope's viewing telescope, and projecting into an f=50mm camera, using a magnification of 24:
dn/dë=(1.76197-1.74805)/ | 4358.35-4799.9107 |=3.1524545*10

^{-5}/A.dE/dn=2/cos{arcsin{n

_{4358.35}/2}}=4.22704rad.dE/dë=4.22704rad*3.1524545*10

^{-5}/A=1.3325551**10^{-4}rad/A.Now using the viewing telescope amounts to multiplying the spectrum's angular width ÄE, by M the magnification. And since f=50 for the projecting lens, we get:

ds/dë=50*24*1.3325551*10

^{-4}rad/A=0.16mm/A.The last value is equivalent to dë/ds=6.25A/mm. This means that our scale must have roughly 6.25A for every mm on film. Upon enlargement of say 5 times, the actual value we get is 1.25A/mm.

We should know roughly which area we are photographing. Otherwise we cannot create scales to use on photographs.

CAUTION!: The dispersion of light on any prism is *non-linear*. This means that the scale we create will be valid only around a small neighborhood of the central wavelength on the picture we take. I.e., it doesn't make sense to photograph an area of 1000 Angstroms width, and place a linear scale under it with subdivisions of 150 A/mm. The greater the magnification, the more it resembles a linear scale.

The quantity dE/dn=2/cos{arcsin{n_{4358.35}/2}} is valid only if the prisms are set to the position of minimum deviation for the wavelength 4358.35A. Since on the Phasmatron spectroscope we can scan other areas too, a more precise calculation must be introduced, namely one that picks the dE/dn as ÄE/Än from the formulas on the section on Spectrum Angular Width. In particular, different incidence angles give rise to different spectrum widths. Thus, they must be taken into account. Then,

We pick the value of ÄE/Äë from running the program with two different input angles, say 68.18 which gives 4358.55811A and 68.19 which gives 4357.82520A. And then,

dE/dë={(68.19-68.18)*(ð/180)}/(4358.55811-4357.8252)=2.38138*10^{-4} rad/A.

This finally gives at magnification 24x, with a 50mm camera lens:

ds/dë=50*24*2.38138*10^{-4}rad/A=0.2857656mm/A or dë/ds=3.5A/mm.

Upon enlarging say 5 times, we get 0.7 A/mm.

We should use the last method, because the program itself takes into account the different angle functions, thus its output correctly reflects the prism positions.