A transparent diffraction grating of a quartz spectrograph is `25mm` wide and has `250` lines per millimetre. The focal length of an objective in whose focal plane a photographic plate is located is equal to `80cm`. Light falls on the grating at right angles. The spectrum under investigating includes a doublet with components of wavelength `310.154` and `310.184nm`. Determine:
(a) the disatnce on the photographic plate between the components of this doublet in the spectra of the first and the second order,
(b) whether these components will be resolved in these orders of the spectrum.

Explanation

From `d sin theta = k lambda`
we get `delta theta = (k delta lambda)/(d cos theta)`
On the other hand `x = fsin theta`
so `delta x = f cos theta delta theta = (kf)/(d) delta lambda`
For `f = 0.80m, delta lambda = 0.30nm` and
`d = (1)/(250)mm`
we get `delta x= {(6,mu,m,if,k,=,1),(12,mu,m,if,k,=,2):}`
(b) Here `N = 25 xx 250 = 6250`
and `(lambda)/(delta lambda) = (310.169)/(0.03) = 10339.. gtN`
and so resolve we need `k = 2` For `k = 1` givens an `R.P.` of only `6250`.