Question

A grating with 400 lines per mm is illuminated with light of wavelength 600.0 nm. a Determine the angles at which maxima are observed b Determine the largest order that can be seen with this grating and this wavelength

Answer 1

Answer:

(a) angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4

Explanation:

(a) for diffraction maxima,

$sin \theta =m\times \lambda/d$

Here, m is the order, $\lambda$ is the wavelength, $\theta$  is the angle at which maxima occur, d is inter planar spacing.

And we know that lines per mm (N) is related with d as,

$N=\frac{1}{d}$

Given that the wavelength is,

$\lambda=600.0 nm=600\times 10^{-9}m$

And $N=\frac{400 lines}{mm} \\N=\frac{400 lines}{10^{-3}m }$

Now,

$sin \theta =m\times \lambda\times N$

Therefore,

$sin \theta= m\times600\times 10^{-9} \times 400\times 10^{3}\\sin \theta=0.24m$

Here, m can be 1,2,3,4 as sin theta has to be less than 1.

$\theta = arcsin 0.24 , arcsin 0.48 , arcsin 0.72 , arcsin 0.96$

Therefore, angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side

b] largest order = 4