Graduate Understanding Modified Bragg's Law with Derivation and Examples

  • Thread starter Thread starter saybrook1
  • Start date Start date
  • Tags Tags
    Bragg's law Law
Click For Summary
The discussion focuses on the derivation and understanding of a modified version of Bragg's Law, which combines elements of Snell's Law and Bragg's Law. Participants express difficulty in deriving the equation and seek clarification on the corrections involved, particularly regarding the refractive index and angle measurements. A key point of contention is the sign of the correction factor, with some arguing for a positive correction while others suggest a negative one based on different interpretations of the refractive index. The conversation highlights the importance of geometry in these calculations and the need for consensus on the correct formulation. Overall, the thread emphasizes collaborative problem-solving in understanding complex physics concepts.
saybrook1
Messages
101
Reaction score
4
Hi guys, the x-ray data booklet gives a modified Bragg's Law that seems to be a combination of Snell's and Braggs. I'll post a picture of what this looks like. I've tried combining the two equations and coming up with their answer but can't get a solid derivation. Any help or a point toward a derivation would be awesome. Thanks!

http://imgur.com/a/CvUGz

http://imgur.com/a/CvUGz
 
Physics news on Phys.org
saybrook1 said:
Hi guys, the x-ray data booklet gives a modified Bragg's Law that seems to be a combination of Snell's and Braggs. I'll post a picture of what this looks like. I've tried combining the two equations and coming up with their answer but can't get a solid derivation. Any help or a point toward a derivation would be awesome. Thanks!

http://imgur.com/a/CvUGz

[PLAIN]http://imgur.com/a/CvUGz[/QUOTE]

I would really like to know where this correction comes from.
 
Last edited by a moderator:
I get something similar, but a couple of corrections. I don't know if my calculations are correct, but I can show you what I got.
One problem with the Bragg equation is the ## \theta ## is not measured from the normal to the surface. In the following derivation, I will use ## \theta ## as from the normal, and ## \theta ' ## as measured from the surface. The index "n" is assumed to be approximately 1 but is assumed to be ## n=1+\delta ##. (This is one of two places where I don't agree completely with what they stated.) ## \\ ## Beginning with ## 2nd cos(\theta_r)=m \lambda ## for constructive interference, and using Snell's law ## n sin(\theta_r)=sin(\theta_i) ##, then ## sin(\theta_r)=sin(\theta_i)/n ##. Also ## sin(\theta_i ')=cos(\theta_i) ## which will be used momentarily. We have ## cos(\theta_r)=(1-(sin(\theta_i)/n)^2)^{1/2} ##so that ## n cos(\theta_r)=(n^2-sin^2(\theta_i))^{1/2}=(n^2-1+1-sin^2(\theta_i))^{1/2}=(n^2-1+cos^2(\theta_i))^{1/2}=(n^2-1+sin^2(\theta_i '))^{1/2} ##Now expand with ## n^2-1=2 \delta ## (approximately)and ## sin(\theta_i ') ## being the larger term. This gives ## n cos(\theta_r)=sin(\theta_i ')(1+2 \delta/sin^2(\theta_i))^{1/2}=sin(\theta_i ')(1+\delta/sin^2(\theta_i ') ) ##. Now we have that ## 2d sin(\theta_i ')=m \lambda ## (Bragg's law without correction).So that ## sin^2(\theta_i ')=(m \lambda)^2/(4 d^2) ## . Putting it all together: ## \\ ## $$ 2d sin(\theta_i ')(1+4 d^2 \delta/(m \lambda)^2)=m \lambda $$. I will try to proofread my response carefully, but I think I have done it correctly. Note that I get a "+" sign for the correction part, not in concurrence with the attachment in the OP.
 
Last edited:
  • Like
Likes saybrook1
Charles Link said:
I get something similar, but a couple of corrections. I don't know if my calculations are correct, but I can show you what I got.
One problem with the Bragg equation is the ## \theta ## is not measured from the normal to the surface. In the following derivation, I will use ## \theta ## as from the normal, and ## \theta ' ## as measured from the surface. The index "n" is assumed to be approximately 1 but is assumed to be ## n=1+\delta ##. (This is one of two places where I don't agree completely with what they stated.) ## \\ ## Beginning with ## 2nd cos(\theta_r)=m \lambda ## for constructive interference, and using Snell's law ## n sin(\theta_r)=sin(\theta_i) ##, then ## sin(\theta_r)=sin(\theta_i)/n ##. Also ## sin(\theta_i ')=cos(\theta_i) ##. This gives ## cos(\theta_r)=(1-(sin(\theta_i)/n)^2)^{1/2} ##so that ## n cos(\theta_r)=(n^2-sin^2(\theta_i))^{1/2}=(n^2-1+1-sin^2(\theta_i))^{1/2}=(n^2-1+cos^2(\theta_i))^{1/2}=(n^2-1+sin^2(\theta_i '))^{1/2} ##Now expand with ## n^2-1=2 \delta ## (approximately)and ## sin(\theta_i ') ## being the larger term. This gives ## n cos(\theta_r)=sin(\theta_i ')(1+2 \delta/sin^2(\theta_i))^{1/2}=sin(\theta_i ')(1+\delta/sin^2(\theta_i ') ) ##. Now we have that ## 2d sin(\theta_i ')=m \lambda ## (Bragg's law without correction).So that ## sin^2(\theta_i ')=(m \lambda)^2/(4 d^2) ## . Putting it all together: ## \\ ## $$ 2d sin(\theta_i ')(1+4 d^2 \delta/(m \lambda)^2)=m \lambda $$. I will try to proofread my response carefully, but I think I have done it correctly. Note that I get a "+" sign for the correction part, not in concurrence with the attachment in the OP.
Oh wow, thanks a ton! This looks great.

So far I had this:

http://imgur.com/a/xJsHR

Lol not even close. This was my latest attempt at least. Tried a ton of different ways to make sense of this. I didn't consider the geometry enough.

Also, from your first form of the Bragg eqn, what happened to the 'n' term? Clearly, it's not part of the answer although I don't see where it disappears.
 
Last edited:
I edited it just a moment ago, (a minor change), but I might continue to update it if I see any additional typos, etc., so please look at my original post once more, etc.
 
  • Like
Likes saybrook1
Charles Link said:
I edited it just a moment ago, (a minor change), but I might continue to update it if I see any additional typos, etc., so please look at my original post once more, etc.

Cool, will do; Thanks again man!
 
  • Like
Likes Charles Link
saybrook1 said:
Oh wow, thanks a ton! This looks great.

So far I had this:

http://imgur.com/a/xJsHR

Lol not even close. This was my latest attempt at least. Tried a ton of different ways to make sense of this. I didn't consider the geometry enough.

Also, from your first form of the Bragg eqn, what happened to the 'n' term? Clearly, it's not part of the answer although I don't see where it disappears.
The "n " term is multiplying ## cos(\theta_r) ##. It multiplied the parenthesis of ## (1-sin^2(\theta_i)/n^2)^{1/2} ## to give ## (n^2-sin^2(\theta_i))^{1/2} ##.
 
  • Like
Likes saybrook1
Charles Link said:
The "n " term is multiplying ## cos(\theta_r) ##. It multiplied the parenthesis of ## (1-sin^2(\theta_i)/n^2)^{1/2} ## to give ## (n^2-sin^2(\theta_i))^{1/2} ##.

I see. Great! I'll try to talk to someone about the sign discrepancy... it's listed other places with the negative sign as well.
 
I think the sign error comes from the original form of Braggs law used. I think if we start with $$m\lambda=2nd\sin(\theta_r)$$ then we can remedy the sign error.
 
  • #10
saybrook1 said:
I see. Great! I'll try to talk to someone about the sign discrepancy... it's listed other places with the negative sign as well.
My equations assume a constructive interference between each of the atomic layers throughout the material. I think I did it correctly. If I got a wrong sign for some reason, it wouldn't be the first time. I'm assuming a positive correction ## \delta ## for the refractive index...
 
  • #11
Charles Link said:
My equations assume a constructive interference between each of the atomic layers throughout the material. I think I did it correctly. If I got a wrong sign for some reason, it wouldn't be the first time. I'm assuming a positive correction ## \delta ## for the refractive index...

Ahhh okay, fair enough.
 
  • #12
saybrook1 said:
Ahhh okay, fair enough.
A google just now, I think, supplies the answer. The article stated, in talking about x-rays, that the index of refraction is just slightly less than 1. Thereby they are using a positive ## \delta ## in your textbook, but use the definition ## n=1-\delta ##. Looks like we are now in concurrence with the textbook result. :-) :-)
 
  • Like
Likes saybrook1
  • #13
Charles Link said:
A google just now, I think, supplies the answer. The article stated, in talking about x-rays, that the index of refraction is just slightly less than 1. Thereby they are using a positive ## \delta ## in your textbook, but use the definition ## n=1-\delta ##. Looks like we are now in concurrence with the textbook result. :-) :-)
Beautiful, so then we can say $$n^2-1\approx-2\delta$$ Do I have that right? Then we'll get the negative sign in the expansion. Thank you.
 
  • Like
Likes Charles Link

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
10K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K