Question

calculate the minimum number of lines that a grating must have to resolve sodium doublet(5890A and 5896A) in third order.

Answer 1

Resolving power of grating is given by:

$\frac{\lambda }{\Delta \lambda }=nN$

$n=order\:of\:grating$

$N=number\:of\:lines\:of\:grating$

$Here\:n=3\:\:\:$

$N=\frac{\lambda }{3\Delta \lambda }$

$\Delta \lambda =5896-5890=6$

$N=\frac{\left(\frac{5896+5890}{2}\right)}{3\times 6}$

$N=\frac{5893}{18}$

$N=327$

Minimum number of lines$=$ 327