Knowledge Bank

The true intensity curve could be measured if one had a perfect spectrometer with no diffraction and used infinitely narrow slits. However, in practice a real spectrometer always detects radiation of other frequencies than the one at which it is set. The relationship between the quantity the spectrograph measures and the true intensity curve can be expressed as the following integral: \begin{equation}{T}(\nu^{\prime})=\int{I}(\nu){g}(\nu, \nu^{\prime}){d}\nu\end{equation} where T is the measured intensity curve determined as a function of $\nu \prime$, the spectrometer setting; I($\nu$) is the true intensity curve; and $g(\nu ,\nu \prime )$ is the slit function. The quantity $g(\nu ,\nu \prime )$ gives the amount of light of frequency $\nu$ striking the sensitive element when the spectrometer is set at $\nu \prime$. The problem rests with the solution of the above integral equation. This of course presupposes a knowledge of g. In this connection, assuming a ""triangular"" slit function and a gaussion distribution of intensity for an emission line, Brodersen has derived the following approximate equation: {L}^{2}=\sigma^{2}+{S}^{2} where L is the half-maximum width of the spectrometer tracing, is the true half-maximum width of the emission line, and s is the effective spectral slit width. The slit function (when normalized) used in the above derivation was of the form: \begin{eqnarray*}{g}(\nu, \nu^{\prime})=\frac{1}{{S}}\left{1-\frac{|\nu-\nu^{\prime}|}{{S}}\right};|\nu-\nu^{\prime}|\leq{S}\=0;|\nu-\nu^{\prime}|>{S}\end{eqnarray*} If s is proportional to the mechanical slit width then the equation, \begin{equation}{L}^{2}=\alpha^{2}(\Delta\eta)^{2}+\sigma^{2}\end{equation} should hold, where $\Delta \eta$ is the mechanical slit width. The quantity $\alpha$ has the form $1/\lambda f(d\nu /d\xi )$ where $\lambda =d\theta /d\xi$ f is roughly four times the focal length of the collimator in the spectrograph (for double pass operation), $\theta -\theta 0$ is the wavelength drum reading, and $\xi$ is the difference between the angles of the Littrow mirror for some arbitrary $\xi$ and when $\xi$ is zero. Measurements have been made on one Hg emission line and two HBr absorption lines. Assuming equation (2) holds for absorption lines also, Table I gives values of $\nu ,\alpha ,\lambda ,d\nu d\xi$ and f. [FIGURE] Several mathematical methods have been tried for the solution of equation (1) using a triangular slit function. The results will be discussed.