Has anyone here used a laser to do a double slit experiment?

Ah! Double Slit experiment! *digs out physics notebook* I think I can pull out an equation to determine wavelength!


ƛ=(dy)÷L


ƛ is wavelength of your laser

d is distance between the centers of the two slits

y is how far the center of the first bright spot from the middle is from the center of the pattern

L is the distance from the slits to the wall the pattern's on

I feel like this is right... Anybody want to confirm this?

spudman2 said:

ƛ=(dy)÷L


ƛ is wavelength of your laser

d is distance between the centers of the two slits

y is how far the center of the first bright spot from the middle is from the center of the pattern

L is the distance from the slits to the wall the pattern's on

I feel like this is right... Anybody want to confirm this?

Click to expand...

Yup, that's the equation for two consecutive bright spots (I think they are called fringes in English)

So... If I had a diffraction grating, I could use a laser of known wavelength to find the gratings lines/cm?

spudman2 said:

So... If I had a diffraction grating, I could use a laser of known wavelength to find the gratings lines/cm?

Click to expand...

You mean the width of the slits?

Here's an experiment that blows my little noggin:

https://www.youtube.com/watch?v=hSRTvKgAs9c

Here's another: https://www.youtube.com/watch?v=E7Xjr-Cdu5M

Here's a Quantum Eraser Demonstration Kit: http://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=6957
.
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Constandinos97 said:

You mean the width of the slits?

Click to expand...

Yes. I suppose I'd just solve for d in the above equation.

spudman2 said:

Yes. I suppose I'd just solve for d in the above equation.

Click to expand...

d is the distance between the two slits, not their width. For the interference pattern to appear, the size of the slits must be comparable to the wavelength of the waves emitted by the laser. So, figuring out their width sounds kind of nonsensical to me (since you can already see the diffraction of the waves)...

I mean, if you're using a laser of about 532nm, then the width of the slits should be something similar, like 500nm (assuming the interference pattern is observed). Now of course, calculating their exact width is harder. I'm not sure that I know which equations are needed for this.

I speculate (best I can do) that knowing the distance between the fringes when using a slit of the exact same size of the wavelength of the laser and the size of the slits when you don't get an interference pattern would help in this case... I'm not quite sure though

I've been pondering the delayed choice double slit experiment for awhile now, this morning it occurred to me that perhaps the reason this works isn't because anything changes with the photon, but rather, we change our frame of reality.

Constandinos97 said:

For the interference pattern to appear, the size of the slits must be comparable to the wavelength of the waves emitted by the laser.

Click to expand...

You mean something like the wavelength is too big, and the wave can't fit through the slit?

spudman2 said:

You mean something like the wavelength is too big, and the wave can't fit through the slit?

Click to expand...

I'm not certain of what would happen. I think the wave would still pass through, but the diffraction of it surely wouldn't occur

Constandinos97 said:

I'm not certain of what would happen. I think the wave would still pass through, but the diffraction of it surely wouldn't occur

Click to expand...

I don't see why it wouldn't, although I will admit I've only seen diffraction demonstrations with 50um? slit width.

spudman2 said:

I don't see why it wouldn't, although I will admit I've only seen diffraction demonstrations with 50um? slit width.

Click to expand...

Oh you mean μm, micrometer. Are you asking why it wouldn't pass or why the diffraction wouldn't occur?
I think it would pass but the wave wouldn't diffract. For example, sound waves have relatively large wavelengths. If you scream through a tiny hole you can still hear the sound on the other side. Of course sound isn't a transverse wave, but it doesn't matter.
Now, arguing whether the tiny hole's size is still comparable to the wavelength of the sound wave is another story...