Brags laww

Physics Q&A
1

Dear course team,

In the lecture notes of lecture notes of lecture 10 braggs law is derived as \sin ( \theta ) =\frac{\lamba}{2d}
however in the solution of excersice 3.4 it is stated that brags law is \sin( 2 \theta ) =\frac{lamba}{d} is there a mistake somewhere or am i missing something.

my confusion comes from the fact that i dont know why we use 2 \theta for powder scattering.

regards Luka

2

Hi Luka!
The use of 2\theta is a mix of convention and how the experimental setup looks like. You have an example in this image, and you may read more about it here.
imagen
In any case, all you need to know is that in Bragg’s law what comes in is \theta, but for experimental reasons what is recorded is the value of 2\theta.
Let me know if that clarifies!