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 June 16th, 2010, 12:53 AM #1 Member   Joined: Aug 2008 Posts: 44 Thanks: 0 Sine? I'm (most likely) missing something ... picture: Shouldn't $\displaystyle \sin\theta=\displaystyle{\frac{d}{\frac{\Delta}{2} }}$? Last edited by skipjack; February 28th, 2018 at 02:33 PM.
 June 16th, 2010, 07:45 AM #2 Member   Joined: Jun 2010 Posts: 36 Thanks: 0 Re: Sine? ? is the angle of incidence of x-ray on the crystal with respect to the surface of the crystal. To find the path difference, between two rays, you have drawn a perpendicular from one ray to the other ray. Let A be the top layer and D be the bottom layer of the crystal. Let AB and AC be the normals to the other ray. From the figure you can see that angle BAD is equal to ?. ?D = the path difference = ?/2 = ADsin? = dsin?.
 June 16th, 2010, 08:47 AM #3 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Sine?
June 17th, 2010, 03:50 AM   #4
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Quote:
 Originally Posted by sa-ri-ga-ma Let AB and AC be the normals to the other ray. From the figure you can see that angle BAD is equal to ?. ?D = the path difference = ?/2 = ADsin? = dsin?.
But the path difference, namely ?/2, misses one part above dashed line in aswoods' figure. That is, it misses $\displaystyle AB*\tan(90-\theta)$, doesn't it?

PS: Aswoods, I first thought this was from some other textbook!

Last edited by skipjack; February 28th, 2018 at 02:36 PM.

 June 17th, 2010, 05:51 AM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Sine? The wave front is not horizontal. So the tiny segment between the top of the dashed line and the upper horizontal line is not part of the path difference.

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