B No intensity drop if slit width is negligible in double slit

[ycheng18]

In our textbook, it says that the intensity of a double slit interference pattern stays the same if the slit width is negligible. I do not understand this concept.

As far as I am concerned, intensity should decrease as the circle of diffraction increase, as intensity is work per area, and the area is increasing as the circle increase but the amount of energy is not. When the interference pattern reaches a far spot, wouldn't the intensity of both waves drop to a low amount, and therefore by adding them up it is still not as bright?

P.S. I am still in high school, so, if possible, try to explain it with high school level material

 

[jtbell]

In our textbook, it says that the intensity of a double slit interference pattern stays the same if the slit width is negligible. I do not understand this concept.

Either you are misinterpreting what the book says, or some important context is missing, or the book is wrong. Can you show us what the book says, exactly?

 

[BvU]

[edit] sorry, I went the wrong way.

Negligible means $width << \lambda$ and there is no single slit diffraction pattern (it is so wide you don't see much difference in intensity from one peeak to another)

Not negligible means width and $\lambda$ are of the same order of magnitude and you see a (wide) envelope from the single slit diffraction pattern.

When the slit width is no longer negligible, that doesn't mean a wider slit, but a narrower slit. In that case the slit is so narrow that the wavelength is no longer << slit width and you see the effect of single slit diffraction.

 

[ycheng18]

Either you are misinterpreting what the book says, or some important context is missing, or the book is wrong. Can you show us what the book says, exactly?

This is from the Tsokos Textbook. The textbook also says the width will never be negligible, but I am nonetheless interested in why it would work like that.

 

[ycheng18]

When the slit width is no longer negligible, that doesn't mean a wider slit, but a narrower slit. In that case the slit is so narrow that the wavelength is no longer << slit width and you see the effect of single slit diffraction.

Thank you for the link of single slit diffraction. However, even if it does exhibit the properties of single slit experiment, the question is still not answered. In a single slit, the intensity in the middle is brighter, and the intensity falls as you move away from the center. The textbook is, however, saying that no matter how far away you move from the center, in a double slit with "negligible length", the intensity of the fringes will still be the same.

 

[BvU]

The intensity from peak to peak will be more the same if the slit width is << $\lambda$ . The limiting case is infinitely wide (zero slit width) but only in theory.
Of course with wider slits the absolute intensity is bigger. But the intensity drops off from peak to peak.

 

[ycheng18]

The intensity from peak to peak will be more the same if the slit width is << $\lambda$ .

I think this may have answered my question. Can you expand on that and hopefully use more logic, less equations (I am trying to understand why)?

Thank you.

 

[BvU]

As far as I am concerned, intensity should decrease as the circle of diffraction increase, as intensity is work per area, and the area is increasing as the circle increase but the amount of energy is not.

You are correct. The criterion for constructive interference ("all the light adds up") from many slits at distances d from each other is that $d\sin\theta=n\lambda$. The pattern drawn is only correct as long as $\sin\theta\approx \theta\approx\tan\theta$. When you go further out, that's no longer true. On a flat screen you also have your $1/r$ effect. And on top of all that you always have the single slit diffraction pattern. It's more a matter of proportions than absoluteness.

 

[ycheng18]

You are correct. The criterion for constructive interference ("all the light adds up") from many slits at distances d from each other is that $d\sin\theta=n\lambda$. The pattern drawn is only correct as long as $\sin\theta\approx \theta\approx\tan\theta$. When you go further out, that's no longer true. On a flat screen you also have your $1/r$ effect. And on top of all that you always have the single slit diffraction pattern. It's more a matter of proportions than absoluteness.

Are you saying that the intensity didn't fall (refer to my post with images earlier) in a "negligible width" only on a very small scale? That when I look at the sides of an interference pattern (where sin⁡θ≈θ≈tan⁡θ is not true), the interference pattern does, in fact, dim?

 

[BvU]

I think this may have answered my question. Can you expand on that and hopefully use more logic, less equations (I am trying to understand why)?

Thank you.

A bit hard. Perhaps this link ?

When the slit width is $<<\lambda$ all the openings function more like point sources.
With bigger slit width, they are extended sources and in some directions the contributions from parts of the source interfere with each other in such a way that the intensity becomes zero.

 

[BvU]

Are you saying that the intensity didn't fall (refer to my post with images earlier) in a "negligible width" only on a very small scale? That when I look at the sides of an interference pattern (where sin⁡θ≈θ≈tan⁡θ is not true), the interference pattern does, in fact, dim?

Yes. If the screen where the fringes are observed is flat, your distance effect then reduces intensity further out. I'm not completely sure about a screen that is shaped as a cylinder with the slits on the axis. Worry about that later is my lazy advice

The picture in your book is a bit misleading in that it shows the same intensity for the central peak. That (and not thinking clearly) is what wrong-footed me in my first reply.

 

[jtbell]

In our textbook, it says that the intensity of a double slit interference pattern stays the same if the slit width is negligible. I do not understand this concept.

Either you are misinterpreting what the book says, or some important context is missing, or the book is wrong.

OK, now I see that I misunderstood what you wrote. I thought you said that the overall intensity (height) of the pattern stays the same as the slit width decreases, but you were actually referring to the fact that for a given slit width, all the peaks have the same height if the slit width is negligible, whereas the peak heights decrease as you go away from the center of the pattern if the slit width is not negligible.

In general, the interference pattern for two slits is a "superposition" or "combination" of two patterns:

1. the pattern that you get when you have two slits with negligble width ("double slit interference")
2. the pattern that you get with a single slit with non-negligible width ("single slit diffraction")

See the "double slit diffraction" diagram on this page:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/mulslid.html#c2

As the slit width decreases, the width of the "single slit envelope" increases. When the slit width is very very small, the "single slit envelope" spreads out so much that all you "see" is a small part of the central maximum, which is almost uniform in height.