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

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

b] largest order = 4

Step-by-step explanation:

(a) for diffraction maxima,

`sin \theta =m* \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=(1)/(d)`

Given that the wavelength is,

`\lambda=600.0 nm=600* 10^(-9)m`

And
`N=(400 lines)/(mm) \\N=(400 lines)/(10^(-3)m )`

Now,

`sin \theta =m* \lambda* N`

Therefore,

`sin \theta= m*600* 10^(-9) * 400* 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