Quantum aspect of light waves with connection with Youngs Double Slit Experiment

[dc5itr888]

So let's say for example there is a slide with two slits on it infront of a screen. A beam of electrons is aimed at it.

A) What would happen to the size of the fringes on the screen if I decreased the spacing between the slits

B) What would happen to the fringes if I moved the screen closer to the slide containing the slits.

C) Would it change if the object being omitted in this case electrons is changed to wave and light waves?


For A, I think by decreasing the spacing between the slits, the width of the fringes on the screen would be more narrow and thinner compared to if it was larger at its original distance.

For B, I think the fringes would be wider and thicker compared at the screens original distance.

For C) I think for both water and light waves going thru youngs double slit experiment would be the same

Can anyone confirm if this explanation is correct? Thanks

 

[dc5itr888]

anyone?

 

[Alter Baron]

First, remember how diffraction works:

- The longer the wavelength, the more diffraction.
- The shorter the wavelength, the less diffraction.

How does the wavelength of the electrons compare with the wavelength of photons?
How does the wavelength of water waves compare with the wavelength of photons and electrons?

What would more diffraction look like? It would mean the light would spread out more as it passes through the slits.

Now, let's look at interference. To be absolutely precise, use the formula (for constructive interference):

d sin(theta) = n(lambda)

where d is the distance between the two slits, theta is the angle between the central maximum and the nth maximum, n is an integer (corresponding to nth-order maximum) and lambda is the wavelength of the incident light on the slits.

Larger theta = larger fringes. Watch how theta changes in each of your scenarios (or does it change?)

To answer the question on the size of the fringes as you move the screen, use the approximation for sin (theta):

sin(theta) = tan(theta) = adj/opp

this approximation holds for small theta. The adjacent side of the triangle in question is the distance from the slits to the screen (remember how theta was defined). The opposite side is the distance between the central maximum and the nth-order maximum, i.e. the size of the fringes.

 

[dc5itr888]

After taking what your said into consideration... I have changed my answer


by decreasing the spacing between the slits, the fringes would be wider and thicker

if I moved the screen closer to the slide containing the slits, fringes on the screen would be more narrow and thinner


Does that sound right? Can anyone confirm?

 

[Alter Baron]

That's correct.