Diffraction grating and wavelength

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CAH
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Hi I've attached the question and mark scheme,

I don't understand part 6(d)... Why have they used sin90?

nλ=dsinθ

If there is another order beam how can it be known the angel at which it exists from the zero order?
Thanks
 
Attachments
  • Diagram of a spectrometer: light from a source enters a circular housing and hits a diffraction grating; a straight-through line A is shown, with the diffracted beam at angle θ to it, captured by a telescope viewed by an observer, labelled B.
    image.jpg
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  • Exam question page showing part 6(c), asking for the wavelength given a diffraction angle of 51.0 degrees and 1480 lines per mm, with a blank answer line in metres, and part 6(d), asking whether more orders could be observed at that wavelength, worth 2 marks.
    image.jpg
    23.5 KB · Views: 565
  • Marking scheme table for question 6d on a diffraction grating: n = d(sin 90)/λ or n = 6.757 x 10^-7 / 5.25 x 10^-7, giving 1.29 so no more beams observed; alternative sin θ = 1.55 so no more beams.
    image.jpg
    24.5 KB · Views: 582
CAH said:
Why have they used sin90?
That is the largest angle where you can observe anything and it maximizes the sine. Either something is observable at this or a smaller angle, or it is not observable at all.
CAH said:
If there is another order beam how can it be known the angel at which it exists from the zero order?
The 0th order is always at 0 degrees.