The ultimate resolving power `lambda//delta lambda` of the spectrograph's trihedral prism is determined by diffraction of light at the prism edges (as in the case of a slit). When the prism is oriented to the least deviation angle in accordance with Rayleigh's criterion,
`lambda//delta lambda = b|dn//d lambda|,`
where `b` is the width of the prism's base (Fig.) and `dn//d lambda` is the dispersion of its material. Derive this formula.

Explanation

Suppose the incident light consists of two wavelength `lambda` and `lambda+ delta lambda` which are just resolved by the prism. Then by Rayleight's criterion, the maximum of the line of wavelength `lambda` must coincide with the first minimum of the line of wavelength `lambda + delta lambda`. Let us write both conditions in terms of the opticak path diferences for the exterme rays:
For the light of wavelength `lambda`
`bn - (DC + CE) = 0`
For the light of wavelength `lambda + delta lambda`
`b(m + deltan) - (DC +CE) = lambda + delta lambda`
beacuse the path difference bwteen extreme rays equal `lambda` for the first minimum in a single slit diffraction (from the fromula `a sin theta = lambda`).
Hence `b dleta n ~~lambda`
and `R = (lambda)/(delta lambda) = B |(deltan)/(delta lambda)| = b |(dn)/(d lambda)|`