Why Must the Slit Size Be Comparable to Wavelength for Diffraction?

[C_Ovidiu]

Can someone please explain this ?
Let's say u have a slit through which waves have to pass .
In order to have a difraction why does the slit have to be comparable with the wavelenghth ? It states this in every book and says that it's demonstrable using Huygens principle . I know Huygens principle but I can't see why should there be a difraction at all , the front wave being parallel with the slit all the time .
Help

 

[Claude Bile]

The slit does not have to be comparable to wavelength, in fact, there does not even need to be a slit! The thing that causes waves to diffract is the fact that they are truncated or "cut-off" at some point. For an infinite plane wave, all the Huygens' wavelets sum to give an infinite planar wavefront, however if we truncate the wave using a slit or by some other method, the Huygen's wavelets around where the wavefront has been "cut-off" no longer sum to give a plane wave, the wavefront becomes bent, which we interpret as the wave spreading out as it propagates.

In a diffraction experiment, the slit is recommended to be around the size of the wavelength of light you are using because these are the best conditions for diffraction to be observed.

Claude.

 

[alvaros]

Not an answer but 1 more question.
The Huygens'algorithm is previous to Maxwell equations. Can this algorithm be infered from the Maxwell equations ?

 

[Claude Bile]

Huygens' principle can be inferred from the wave equation from field continuity arguments, however Huygens' principle is only applicable to propagating waves and does not predict the existence of evanescent waves.

Claude.

 

[olgranpappy]

Can someone please explain this ?
Let's say u have a slit through which waves have to pass .
In order to have a difraction why does the slit have to be comparable with the wavelenghth ? It states this in every book and says that it's demonstrable using Huygens principle . I know Huygens principle but I can't see why should there be a difraction at all , the front wave being parallel with the slit all the time .
Help

Your first diffraction minimum will be located at an angle \theta where
<br /> d\sin(\theta)=\lambda<br />
where d is the slit size and lambda is the wavelength. Since sine is never bigger than one, if d is less than lambda the condition can never be fulfilled. (the extreme case of this is when d is very much smaller than lambda in which case the wavefronts coming out look sphereical and obviously can't interfere with each other).

On the other hand if d is very very big, then very many diffraction minima (and maxima) will occur within a very small angle and will not be easy to see. (the extreme case of this is when d is very much larger than lambda in which case there is no screen at all, it's "all slit").

so, you need d to be bigger than, but comparable to, the wavelength.

Cheers.