Why Must the Wavelength Be Smaller Than the Interplanar Spacing in Bragg's Law?

[Roula]

Hello everyone
I would like to ask you something related to the investigation of materials depending on Bragg Diffraction (Bragg Law).

It is a prerequisite, that the wavelength of the used radiation must be smaller than the dimension of the sample, by the meaning λ ≤ d , where λ is the wavelength and d is the interplaner spacing .

I know from Bragg law nλ=2dsinθ, that wavelength must be smaller than d in order to apply this law. But i can not understand physically the reason of that.
What is the physical reason that λ ≤ d ?

I am thankful for you all.

Roula

 

[Dr Transport]

simply, waves do not scatter unless the wavelength of incident wave is on the order or smaller than the feature you want to investigate.

if that were the case, you would not be able to listen to the radio in your house because the radio waves would be scattered by your wall thickness...

 

[Roula]

Dear Dr. Transport.
i am thankful for your response.

You are right that the wavelength must be smaller than the dimension to get scattered. And that was my question.

Could you please prove it for me ?
does the proof relate to Huygens–Fresnel principle ?

thanks again.
Roula

 

[Roula]

No answer :(

 

[DrDu]

The wavelength don't has to be smaller than d for Bragg's law to be valid. It's just so that for λ<d the equation has the only solution n=0 and θ=0, i.e. the only possible solution is an unscattered wave.

 

[Dr Transport]

Check Newton's book, scattering theory of particles and waves.