# Wave diagram

1. Jan 24, 2008

### Andyblues

Heya! I'm new in the forum, and here is my first entry.

I don't really understand the diagram for wave difraction. Let me try to draw it...

| | | | | The space between the four three lines is suppose to be the wavelength.
| | | | |
| | | | | The last line is suppose to be a single slit.
| | | | |
| | | | | In my textbook it says something like the space of the slit must be
| | | | comparable to the wavelenght...
| | | |-->
| | | | So, if the wavelengt is smaller than the slit, the wave will pass through
| | | | | the with no change.
| | | | |
| | | | | And if the wavelenth is comparable to the space, then it will be
| | | | | diffracted...
| | | | |

In the case of my fantastic diagram, its clear that it will be diffracted, and the wave will form sort of circle forms when passing through the slit.

But if the diagram is the following...

| The wave will pass with absolute freedom... is it right?
|
| Then, my question is... wasn't the wavelength the space between each
| line?? If so... what does it have to do with the lenght of the same lines???

| | |
| | | :surprised
| | |

|
|
I hope my question is clear enough!

Thanks!

Andrés.

2. Jan 24, 2008

### Staff: Mentor

You always get diffraction around the edges of the slit. If the slit is wide, the effects are noticeable only near the edges of the "shadow" cast by the slit. If the slit is narrow (for the same wavelength), the two regions of "edge diffraction" merge and become the dominant effect.

Here's a sketch that shows the brightness pattern on the screen for different slit widths, for the same wavelength (approximately!).

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• ###### diffraction.gif
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Last edited: Jan 24, 2008
3. Jan 24, 2008

### Claude Bile

Diffraction (what we might define as the bending of wavefronts) occurs when we truncate (cut-off) the wavefront by means of an aperture or obstacle. Even for apertures that are very large compared to the wavelength the wave is still diffracted albeit very slightly. As the aperture gets smaller, diffraction effects get larger (i.e. the wavefronts get bent more).

When the aperture size is equal to roughly half the wavelength, you hit an interesting point where the wave diffracts as much as it possibly can - in other words the wavefronts are nearly circular. Decreasing the aperture size beyond this point does not cause the wavefronts to bend further.

- The wave will not be diffracted provided there are no obstacles present to truncate the wavefront.
- The wavelength remains the same at all times.
- What the length of the wavefront has to do with things is a VERY tricky question, you would need some grasp of Fourier theory to understand its significance.
- Essentially, the "length" of the wavefront as it moves through an aperture determines how many spatial frequency components are formed as the wave passes through the aperture.
-The smaller the aperture, the greater the number of spatial frequencies.
- Having more spatial frequency components causes the wave to diffract more!

Claude.

4. Jan 25, 2008

### Andyblues

Hi, thanks guys .

I'm sorry I didn't ask my question properly! Let me try again with a more pro diagram.

In case A, you can see that the space between the edges of the slit is larger thant the lenth of the lines of the wave. Then, the wave pass through the slit "free".

In case B, the space is smaller (in comparision to the length of the wave).

My question was... if the refracction depends in the wavelength (lambda), and it is equal in both cases (you can see that lambda and lambda prime are almost equal), the difference then must be in the length of the lines of the wave in the diagram.... but what does this lenght represent in "real" life??? because I can't see any diference in the wavelength, and its suppose to be the main reason of diffraction!

And... how can be always diffraction?!!! How is it possible? :surprised

Thanks a lot!!!!

Andrés.

5. Jan 25, 2008

### Andyblues

ups, I forgot the attachment

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• ###### wave diagram.JPG
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6. Jan 25, 2008

### Andy Resnick

The length of your lines would correspond to the spatial extent of the wavefront, not the wavelength. And what your drew is, strictly speaking, physically impossible- a planar wavefront must be infinite in extent.

But, your diagram 'A' is close to how a spatial filter operates- a focused beam impinges on a pinhole, with the diameter of the pinhole close to the diameter of the spot size. What emerges is a near perfect spherical wavefront, which can then be transformed into a cleanly collimated beam.

7. Jan 28, 2008

### Claude Bile

The spatial shape of the wavefront is critical in calculating diffraction effects, far more so than wavelength - wavelength simply gives you a reference scale.

As Andy Resnick said, diagram A is technically physically impossible since the wavefront must be infinite (or a very special type of beam) in order for no spreading/bending of the planar wavefronts to occur. A close approximation however is a standard Gaussian beam passing through a wide aperture at the beam waist. One would expect little, if any change to the wavefront due to the aperture.

Diagram B would then correspond to either a wider beam, (or a smaller aperture), in which case there would be significant change to the wavefront due to the aperture.

Claude.