# A Modified Bragg's Law

1. Oct 2, 2016

### saybrook1

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

2. Oct 2, 2016

### saybrook1

Last edited by a moderator: May 8, 2017
3. Oct 2, 2016

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: Oct 2, 2016
4. Oct 2, 2016

### saybrook1

Oh wow, thanks a ton! This looks great.

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: Oct 2, 2016
5. Oct 2, 2016

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.

6. Oct 2, 2016

### saybrook1

Cool, will do; Thanks again man!

7. Oct 2, 2016

The "n " term is multiplying $cos(\theta_r)$. It multiplied the parenthesis of $(1-sin^2(\theta_i)/n^2)^{1/2}$ to give $(n^2-sin^2(\theta_i))^{1/2}$.

8. Oct 2, 2016

### saybrook1

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.

9. Oct 2, 2016

### saybrook1

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. Oct 2, 2016

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. Oct 2, 2016

### saybrook1

Ahhh okay, fair enough.

12. Oct 2, 2016

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.