Light falls normally on a transparent diffraction grating of width `l = 6.5cm` with `200` lines per millimetre. The spectrum under inverstigation includes a spectral line with `lambda = 670.8 nm` consisting of two components differing by `delta lambda = 0.015nm`. Find:
(a) in what order of the spectrum these components will be resolved,
(b) the least difference of wavelength that can be resolved by this grating in a wavelength region `lambda ~~ 670nm`.

Explanation

We see that
`N = 6.5 xx 10 xx 200 = 13000`
Now to resolve lines with `delta lambda = 0.015nm` and `lambda ~~670.8nm` we must have
`R (670.8)/(0.015) = 44720`
Since `3N lt R lt 4N` one must go to the fourth order to resolve the said components.
(b) we have `d = (1)/(200)mm = 5mu m`
so `sin theta = (k lambda)/(d) = (k xx 0.670)/(5)`
`since |sintheta|le` we must have `kle 7.46`
so `k_(max) = 7~~(d)/(lambda)`
Thus `R_(max) = k_(max)N = 91000~~(Nd)/(lambda) = (l)/(lambda)`
where `l = 6.5cm` is the grating width.
Finally `delta lambda_(min) = (lambda)/(R_(max)) = (670)/(91000) = 007nm = 7pm ~~(lambda^(2))/(l)`.