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clear
//Given
l1=5890*10**-10 //m
l2=5896*10**-10
d=2.0*10**-6 //m
D=2 //m
//Calculation
x=(3*D*(l2-l1))/(2*d)
//Result
printf("\n Spacing between the first maxima of two sodium lines is %0.3f *10**-4 m",x*10**4)
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