The answer?
DISPERSION. I say dispersion.
Ray angle deviation and dispersion through a prism can be determined by tracing a sample ray through the element and using Snell's law at each interface. For the prism shown at right, the indicated angles are given by
{\displaystyle {\begin{aligned}\theta '_{0}&=\,{\text{arcsin}}{\Big (}{\frac {n_{0}}{n_{1}}}\,\sin \theta _{0}{\Big )}\\\theta _{1}&=\alpha -\theta '_{0}\\\theta '_{1}&=\,{\text{arcsin}}{\Big (}{\frac {n_{1}}{n_{2}}}\,\sin \theta _{1}{\Big )}\\\theta _{2}&=\theta '_{1}-\alpha \end{aligned}}}.All angles are positive in the direction shown in the image. For a prism in air {\displaystyle n_{0}=n_{2}\simeq 1}. Defining {\displaystyle n=n_{1}}, the deviation angle {\displaystyle \delta } is given by
{\displaystyle \delta =\theta _{0}+\theta _{2}=\theta _{0}+{\text{arcsin}}{\Big (}n\,\sin {\Big [}\alpha -{\text{arcsin}}{\Big (}{\frac {1}{n}}\,\sin \theta _{0}{\Big )}{\Big ]}{\Big )}-\alpha }If the angle of incidence {\displaystyle \theta _{0}} and prism apex angle {\displaystyle \alpha } are both small, {\displaystyle \sin \theta \approx \theta } and {\displaystyle {\text{arcsin}}x\approx x} if the angles are expressed in radians. This allows the nonlinear equation in the deviation angle {\displaystyle \delta } to be approximated by
{\displaystyle \delta \approx \theta _{0}-\alpha +{\Big (}n\,{\Big [}{\Big (}\alpha -{\frac {1}{n}}\,\theta _{0}{\Big )}{\Big ]}{\Big )}=\theta _{0}-\alpha +n\alpha -\theta _{0}=(n-1)\alpha \ .}The deviation angle depends on wavelength through n, so for a thin prism the deviation angle varies with wavelength according to
{\displaystyle \delta (\lambda )\approx [n(\lambda )-1]\alpha }.