Observationelle Værktøjer

The equations of spectroscopy

The prism

The angles involved for rays passing a prism are connected through Snell's law, which for the angles with the normal to the surfaces $\phi_1, \phi_1'$ and $\phi_2', \phi_2$ states

$$\sin\phi_1' = {n \over n'}\sin\phi_1$$ (1)

and

$$\sin\phi_2 = {n' \over n}\sin\phi_2'$$ (2)

where $n'(\lambda)$ is the refractive index for the glass and $n\sim 1$ is the refractive index of air. The two angles inside the prism are connected through the prism top angle $\alpha$

$$\alpha = \phi_1' + \phi_2'$$ (3)

The change of the direction of the ray exiting the prism with respect to the incoming ray is given by

$$\delta = \phi_1 + \phi_2 - \alpha$$ (4)

The minimum deviation $\delta_m$ is found when the angles $\phi_1 = \phi_2$ in which case

$${n'(\lambda) \over n} = {\sin (0.5(\alpha+\delta_m)) \over \sin (0.5\alpha)}$$ (5)

By measuring $\delta_m$ one can determine $n'(\lambda)$. With a table of $n'(\lambda)$, the angular dispersion $d\delta/d\lambda$ can be found. In the notes we have a case, where we find the dispersion to be 0.6 deg/Å. With a focal length of the camera of f_cam = 500 mm the linear dispersion becomes $dx/d\lambda = 0.0052$ mm/Å.

The grating

The light transmitted from one slit will be diffracted according to the wellknown formula

$$I \propto {\sin^2\beta \over \beta^2}$$ (6)

where

$$\beta = \pi{b \over \lambda}(\sin i + \sin \theta)$$ (7)

The phase shift one ray relative to the other is given by AB+BC, which comes to a phase shift $\delta$

$$\delta = 2\pi{d \over \lambda} (\sin i + \sin \theta)$$ (8)

The electrical field from all slits is now made up of sums over terms

$$E_n = a\exp(i(-kx+\omega t + (n-1)\delta)) = a\exp(i(n-1)\delta)\exp(i(-kx-\omega t))$$ (9)

Summing all terms from the n slits give a total amplitude of

$$A = a{1 - \exp(in\delta)\over 1 - \exp(i\delta)}$$ (10)

The intensity I measured is the square of the electrical field or

$$I = a^2{\sin^2(n\delta/2) \over \sin^2(\delta/2)}$$ (11)

Combining the results from the diffraction of one slit with the sum over the total set of slits one gets

$$I = I_0 \times {\sin^2\beta \over \beta^2} \times {\sin^2(n\gamma) \over \sin^2(\gamma)}$$ (12)

and $\gamma$ is defined by

$$\gamma = \delta/2 = \pi{d \over \lambda}(\sin i + \sin \theta)$$ (13)

The last factor will give maximum contribution when the E_n contributions are in phase, which happens for integers m satisfying $\gamma = m\pi$. This leads to the very basic equation (the grating equation)

$$d(\sin i + \sin \theta) = m\lambda$$ (14)

Now, several things can be derived. The theoretical resolution can be derived, which defines the smallest wavelength difference that can be measured. The argument is to move so far from a given wavelength $\lambda$, where you have constructive interference to a wavelength $\lambda + \delta\lambda$, where you have destructive interference. This gives the answer

$$\delta \lambda = {\lambda \over mn}$$ (15)

One defines the theoretical resolution R of a grating by the ratio

$$R = {\lambda \over \delta\lambda} = mn$$ (16)

The angular dispersion is found by differentiation of the grating equation and gives

$${d\theta \over d\lambda} = {m \over d\cos\theta}$$ (17)

To get a high dispersion, it obviously pays off to work with a grating that can be used in a high order mode m. Finally one can find the distance between subsequent orders m and m+1as

$$\Delta \lambda = \lambda/m$$ (18)