With light falling normally on a transparent diffraction grating `10mm` wide, it was found that the components of the yellow line of sodium `(589.0` and `589.6nm`) are resolved beginning with the fifth order of the spectrum. Evaluate:
(a) the period of this grating,
(b) what must be the width of the grating with the same period for a double `lambda = 460.0nm` whose components differ by `0.13nm` to be resolved in the third order of the spectrum.

Explanation

Here
`R = (lambda)/(delta lambda) = (589.3)/(0.6) = kN = 5N`
so `N = (589.3)/(3) = (10^(-2))/(d)`
`d = (3 xx 10^(-2))/(589.3)m = 0509mm`
(b) To resolve a doublet with `lambda = 460.0nm` and `delta lambda = 0.13nm` in the thrid order we must have
`N = (R )/(3) = (460)/(3 xx 0.13) = 1179`
This means that the grating is
`Nd = 1179 xx 0.0509 = 60.03mm`
wide `= 6cm` wide.