Answer:
[tex]\theta=41.29^{\circ}[/tex]
Explanation:
Given that,
Number of lines, [tex]N=12000\ lines/cm[/tex]
[tex]N=12000\times 10^2\ lines/m[/tex]
Wavelength, [tex]\lambda=550\ nm=550\times 10^{-9}\ m[/tex]
We need to find the angle for the first order maximum, n = 1
[tex]d=\dfrac{1}{N}[/tex]
[tex]d=\dfrac{1}{12000\times 10^2}[/tex]
[tex]d=8.33\times 10^{-7}\ m[/tex]
Using the grating equation as :
[tex]d\ sin\theta=n\lambda[/tex]
[tex]sin\theta=\dfrac{\lambda}{d}[/tex]
[tex]sin\theta=\dfrac{550\times 10^{-9}}{8.33\times 10^{-7}}[/tex]
[tex]\theta=41.29^{\circ}[/tex]
So, the angle for the first order maximum is 41.29 degrees. Hence, this is the required solution.
