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June 16th, 2010, 01:53 AM   #1
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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 03:33 PM.
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June 16th, 2010, 08:45 AM   #2
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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?.
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June 16th, 2010, 09:47 AM   #3
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Re: Sine?

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June 17th, 2010, 04:50 AM   #4
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Re: Sine?

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 03:36 PM.
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June 17th, 2010, 06:51 AM   #5
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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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