A spectrograph's trihedral prism is manufactured from glass whose refractive index varies with wavelength as `n = A + B//lambda^(2)`, where `A` and `B` are constats, with `B` being equal to `0.010mum^(2)`. Making use of the formula from the foregoing problem, find:
(a) how the resolving power of the prism depends on `lambda`, calculate the value of `lambda//delta lambda` in the vicinity of `lambda_(1) = 434nm` and `lambda_(2) = 656nm` if the width of the prism's base is `b = 5.0 cm`,
(b) the width of the prism's base capacble of resolving the yellow doublet of sodium `(589.0` and `589.6nm`).

Explanation

`(lambda)/(delta lambda) = R = b|(dn)/(d lambda)| = 2Bb//lambda^(3)`
For `b = 5cm, B = 0.01 mu m^(2) lambda_(1) = 0.434mum = 5 xx 10^(4)mu m`
`R_(1) = 1.223 xx 10^(4)`
for `lambda_(2) = 0.656mu m`
`R_(2) = 0.3542 xx 10^(4)`
(b) Top resolve the `D`-lines we require
`R = (5893)/(6) = 982`
Thus `982 = (0.02 xx b)/((0.5893)^(3))`
`b = (982 xx (0.5893)^(3))/(0.02)mum = 1.005 xx 10^(4)mu m = 1.005cm`