[dipesh747]

Hi, I'm having some difficulty in deriving the bragg equation. I have looked online and the only source I can find was Wikipedia and the proof appears to be wrong on there.

So far I have

AB = BC = d/sin x , AC = 2d / tan x

Then for AC` I have got two answers:

1. I have formulated a triangle with the adjacent being 2d/tan x (Length of AC), the opposite being 2d (I assumed it was 2d as the other wavelength of light had travelled 2d along the horizontal axis).

So to work out the hypotenuse I used Pythagoras so I got AC`^2 = ((4d^2/tan^2 x) + (4d^2)) - if this is the correct method how would I square root this?

The second way round I tried to work out the answer was use Trig with the adjacent being 2d/tan x

Then I got AC` to be 2dcos x/tan x

Neither of these have got me to the answer to 2dsin x. Can anyone see what I have missed?

[cth]

This seems to be overly complicated. 

Look at http://www.eserc.stonybrook.edu/projectjava/bragg/ where there is a straight forward demonstration, middle of page. No need to use square roots and cos x...

[dipesh747]

I see, my starting position made the whole thing more complicated than it needed to be. Thanks