How Do Slits and Gratings Create Light Patterns You See?

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sparkle123
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When you shine light through a diffraction grating/double slit/single slit, I understand that you see an interference/diffraction pattern on the wall, so that you have:
light source -- grating/slit ---pattern on wall

But if you put a grating/slits between your eyes and a light source, why do you see a pattern on the wall in the direction of the light source? i.e. the configuration
eyes --- slits --- light source --- pattern on wall
as opposed to
pattern on wall/eyes --- slits --- light source

Thanks!
 
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Most likely light sources behind you, or reflection (if I understand your question).
 
How would reflection work?
 
If we go by vector diffraction theory we see that the field behind the aperture is

[tex]\vec{E}(\vec{x})=\vec{E}^{(0)}(\vec{x})+ \vec{E}^{(1)}( \vec{x} )-<br /> \vec{E}_{diff}(\vec{x})[/tex]

Where E^0 and E^1 describes the sources behind the aperture given a perfectly conducting plane at the aperture plane.

I'm sure you just want to cut the chase though. In the region behind the aperture, we see both the reflections of the source AND the diffraction pattern. So you can see, the diffraction pattern is produced both in front, and behind the aperture.
 
Are the diffraction patterns produced in front and behind the aperture the same? (so you see one image)?
Thank you! :)
 
sparkle123 said:
But if you put a grating/slits between your eyes and a light source, why do you see a pattern on the wall in the direction of the light source?

First, where does the light actually end up? (what is the actual "screen" in this situation?)

Second, what kind of sensations would you expect this to produce?
 
sparkle123 said:
Are the diffraction patterns produced in front and behind the aperture the same? (so you see one image)?
Thank you! :)
The diffraction itself will be same, but the image will be different. Since detectors (such as eyes) measure the field's intensity, in the reflection case there are three terms that contribute to intensity as a function of position.
 
jtbell said:
First, where does the light actually end up? (what is the actual "screen" in this situation?)
The light that goes through the slits hits your eyes so your eye is the screen?
So the light is reflected off your eyes?

jtbell said:
Second, what kind of sensations would you expect this to produce?
Uhhh... your eye hurts?

Mindscrape said:
The diffraction itself will be same, but the image will be different. Since detectors (such as eyes) measure the field's intensity, in the reflection case there are three terms that contribute to intensity as a function of position.
But if there are two images, why do you only see one?

Thanks you guys!