calculate the minimum number of lines that a grating must have to resolve sodium doublet(5890A and 5896A) in third order.
Resolving power of grating is given by:
λΔλ=nN\frac{\lambda }{\Delta \lambda }=nNΔλλ=nN
n=order of gratingn=order\:of\:gratingn=orderofgrating
N=number of lines of gratingN=number\:of\:lines\:of\:gratingN=numberoflinesofgrating
Here n=3 Here\:n=3\:\:\:Heren=3
N=λ3ΔλN=\frac{\lambda }{3\Delta \lambda }N=3Δλλ
Δλ=5896−5890=6\Delta \lambda =5896-5890=6Δλ=5896−5890=6
N=(5896+58902)3×6N=\frac{\left(\frac{5896+5890}{2}\right)}{3\times 6}N=3×6(25896+5890)
N=589318N=\frac{5893}{18}N=185893
N=327N=327N=327
Minimum number of lines=== 327
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