Answer:
Explanation:
Given
[tex]N=3680 cm^{-1}[/tex]
therefore slit spacing [tex]d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm[/tex]
since [tex]d\sin \theta =n\lambda [/tex]
for [tex]n=1[/tex]
[tex]d\sin \theta =\lambda [/tex]
Now,at [tex]\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm[/tex]
at [tex]\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm[/tex]
at [tex]\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm[/tex]