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Derivation help please

How might one derive this formula?

nλ=dsinθn \lambda = d \text{sin} \theta

I need to know it and my book isn't very clear at all. :confused:
Ahhh good old Braggs law :smile:
Remind me what it's to do with and I'll post how to derive it.

Just flicking through my physics notes here.
Captain Biggles
Remind me what it's to do with and I'll post how to derive it.

Just flicking through my physics notes here.


It's got to do with diffraction gratings, and the wavelenght of monochromatic light.

I looked at that Bragg's law link and it looks much more advanced than what I'm looking for. I'm just at A level so is there any sort of straightforward (and perhaps technically wrong) diagram + 4 lines of algebra and trig derivation of it out there?
HeatherChandler
I looked at that Bragg's law link and it looks much more advanced than what I'm looking for. I'm just at A level so is there any sort of straightforward (and perhaps technically wrong) diagram + 4 lines of algebra and trig derivation of it out there?


Okay sorry, think about a right angled triangle ABC where AB is the distance between two gaps in a diffraction grating (d) and the angle BAC is theta and BCA is 90 degrees. So now sin theta = opp/hypot = BC/AB and BC must be a whole number of wavelengths n(lamda) :smile:
EierVonSatan
Okay sorry, think about a right angled triangle ABC where AB is the distance between two gaps in a diffraction grating (d) and the angle BAC is theta and BCA is 90 degrees. So now sin theta = opp/hypot = BC/AB and BC must be a whole number of wavelengths n(lamda) :smile:


Oh sham thank you for that. I was just drawing the triangle and I was about to ask why BC must be n lambda and then it just clicked! :biggrin:
HeatherChandler
Oh sham thank you for that. I was just drawing the triangle and I was about to ask why BC must be n lambda and then it just clicked! :biggrin:


That tends to happen when doing proofs! :biggrin:

For me anyways. :redface:

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