Correct statement about coherent sources and interference

[songoku]

Homework Statement

Which of the following statements is correct?
a. coherent sources are not needed to produce interference fringes
b. two coherent light sources do not always produce bright and totally dark fringes on a screen
c. the atoms in a tungsten filament lamp produce coherent light waves
d. the yellow light from a street lamp is coherent light

Homework Equations

not sure

The Attempt at a Solution

(a) is wrong because coherent sources are needed to produce interference fringes
(c) is wrong because the movement of atoms is random hence the light waves produced are not coherent
(d) is wrong (same reason as (c))

So only option (b) is left. But I also think it is wrong because if coherent sources are needed to produce interference pattern, does it also mean that two coherent sources will always produce interference pattern (bright and dark fringe)?

Thanks

 

[haruspex]

coherent sources are needed to produce interference fringes

Are you sure? There is some ambiguity in that statement.

 

[Cutter Ketch]

So only option (b) is left. But I also think it is wrong because if coherent sources are needed to produce interference pattern, does it also mean that two coherent sources will always produce interference pattern (bright and dark fringe)?

B said “totally dark”. Will two coherent sources will ALWAYS produce regions of total darkness (total destructive interference)? For example what if the two sources have different amplitude?

 

[Cutter Ketch]

(c) is wrong because the movement of atoms is random hence the light waves produced are not coherent

C is definitely wrong, and the movement of the sources certainly doesn’t help, but the main reason the light from multiple atoms isn’t coherent is because they emit at different times. Each emission is a completely independent event (ignoring the stimulated emission that can also occur) There is no reason for the phases to be related.

 

[songoku]

Are you sure? There is some ambiguity in that statement.

I am not sure about the ambiguity. Maybe you mean coherent sources are needed to produce observable interference pattern? If two light waves just interfere, no need for the two waves to be coherent because any waves can interfere?

B said “totally dark”. Will two coherent sources will ALWAYS produce regions of total darkness (total destructive interference)? For example what if the two sources have different amplitude?

Ahh I see. If the amplitude is different, the waves won't produce totally dark fringes. What if the statement like this: two coherent light sources do not always produce bright and dark fringes on a screen? Is this statement wrong because coherent light sources will always produce bright and dark fringes?

C is definitely wrong, and the movement of the sources certainly doesn’t help, but the main reason the light from multiple atoms isn’t coherent is because they emit at different times. Each emission is a completely independent event (ignoring the stimulated emission that can also occur) There is no reason for the phases to be related.

Is it possible for the waves to be coherent even though they are emitted at different times?

Thanks

 

[Cutter Ketch]

I am not sure about the ambiguity. Maybe you mean coherent sources are needed to produce observable interference pattern? If two light waves just interfere, no need for the two waves to be coherent because any waves can interfere?Ahh I see. If the amplitude is different, the waves won't produce totally dark fringes. What if the statement like this: two coherent light sources do not always produce bright and dark fringes on a screen? Is this statement wrong because coherent light sources will always produce bright and dark fringes?Is it possible for the waves to be coherent even though they are emitted at different times?

Thanks

Yes, the statement would then be wrong. Two coherent sources will always show some interference. (Well, with the trivial caveats that they overlap in the space in question, they at least partially share a common polarization, and they aren’t traveling exactly the same direction)

RE different times, yes, that is how a laser works. The emissions are spread out over time. However in that case the phases are locked together by a process called stimulated emission. Without some mechanism to force the phases to match, independent events will randomly fill all possible phases = incoherent.

 

[haruspex]

I am not sure about the ambiguity.

When can one source look like two?

 

[haruspex]

If the amplitude is different, the waves won't produce totally dark fringes.

That is far from obvious. Most points on the screen are at different distances from the two sources. Maybe on an infinite screen there are always points such that the two distances, f and g say, satisfy both:

|f-g| is an odd multiple of λ/2
(f/g)² is the ratio of the source intensities.​

 

[Cutter Ketch]

That is far from obvious. Most points on the screen are at different distances from the two sources. Maybe on an infinite screen there are always points such that the two distances, f and g say, satisfy both:

|f-g| is an odd multiple of λ/2
(f/g)² is the ratio of the source intensities.​

? His statement is absolutely correct. If the amplitudes are not equal there is no construction that results in 0 intensity. The minimum amplitude possible is the difference in the two amplitudes.

 

[haruspex]

? His statement is absolutely correct. If the amplitudes are not equal there is no construction that results in 0 intensity. The minimum amplitude possible is the difference in the two amplitudes.

If the amplitudes are not equal at the point on the screen then it will not be totally dark. But that is not the question. The question is whether you can construct two sources, each coherent and at the same wavelength, such that no completely dark fringes appear on the screen. My point is that using different amplitudes for the sources does not necessarily produce that result.

 

[Cutter Ketch]

If the amplitudes are not equal at the point on the screen then it will not be totally dark. But that is not the question. The question is whether you can construct two sources, each coherent and at the same wavelength, such that no completely dark fringes appear on the screen. My point is that using different amplitudes for the sources does not necessarily produce that result.

As the amplitude varies in space, I agree that the amplitudes being unequal at one point in space does not assure that they won’t happen to be equal at another point in space. In fact almost every practical case will have places where the amplitudes are equal say in the diminished edges of beams or diffracted shadows. Heck, there has to be somewhere where they are both zero amplitude (ie where the light doesn’t shine).

I’m sure you agree that his statement holds locally: wherever the amplitudes are not equal there can be no totally destructive interference. However, caveats about all space aside, it is not unusual or difficult to construct a case where they are not equal anywhere in a practical region of interest. In most of the usual interference arrangements if the source amplitudes are sufficiently different no totally dark fringes will be seen in the significant region of interference.

 

[songoku]

When can one source look like two?

When double slit is put in front of the source?

You mean that to produce interference fringes, we don't necessarily need more than one source but can use one source and double slit?

That is far from obvious. Most points on the screen are at different distances from the two sources. Maybe on an infinite screen there are always points such that the two distances, f and g say, satisfy both:

|f-g| is an odd multiple of λ/2
(f/g)² is the ratio of the source intensities.​

Sorry I don't get this.

"Most points on the screen are at different distances from the two sources". Ok yes, only center bright fringes will have same distance from the two sources.

"Maybe on an infinite screen there are always points such that the two distances, f and g say, satisfy both:

|f-g| is an odd multiple of λ/2
(f/g)² is the ratio of the source intensities."​

If the distance from the slit to the screen is finite, I think there are also points that satisfy both conditions, |f-g| is an odd multiple of λ/2 to obtain destructive interference and (f/g)² is the ratio of the source intensities from inverse square law. I don't get the hint why the distance between slit and screen should be infinite and where this hint leads to regarding to the question.

If the amplitudes are not equal at the point on the screen then it will not be totally dark. But that is not the question. The question is whether you can construct two sources, each coherent and at the same wavelength, such that no completely dark fringes appear on the screen. My point is that using different amplitudes for the sources does not necessarily produce that result.

I also don't get this part. Let say we have two coherent sources with different amplitudes. I think we won't get completely dark fringes on the screen. Is it not so?
Maybe I misinterpret your post.

Thanks

 

[Cutter Ketch]

I also don't get this part. Let say we have two coherent sources with different amplitudes. I think we won't get completely dark fringes on the screen. Is it not so?
Maybe I misinterpret your post.

To clarify, I don’t think haruspex is denying that unequal amplitudes can’t add up to dark fringes. I think what he is pointing out is that unequal amplitude right here doesn’t mean the amplitudes are unequal everywhere. The amplitude varies over space so in most constructions, even if the source amplitudes are different, you can probably still find a place where the amplitudes are not neglibly small, the amplitudes happen to be equal, and the phases are destructive, say on the fringes of the stronger beam.

So the statement that two sources “don’t always” make dark fringes is true. However your statement “if the the amplitudes are different they won’t make dark fringes”, while certainly true at the location where the amplitudes are not equal, is not guaranteed to be true everywhere. However, as I pointed out, in most experiments where fringes are being observed, sufficiently different amplitudes will, as a practical matter, assure no dark fringes can appear over the region of interest.

 

[haruspex]

You mean that to produce interference fringes, we don't necessarily need more than one source but can use one source and double slit?

Yes, it depends what the question means by two sources.

why the distance between slit and screen should be infinite

No, an infinite screen at a finite distance. The question must assume an arbitrarily large screen to make sense. Otherwise could always avoid a dark fringe by using a screen to small to include it.

Let say we have two coherent sources with different amplitudes. I think we won't get completely dark fringes on the screen.

If the source amplitudes are different then you can find points on the screen satisfying the two conditions I stated in post #8. Those points will be completely dark.
But what about the simplest possible set-up: two equally bright sources, in phase, equidistant from the screen? Will there be points on the screen satisfying the conditions now?

 

[Cutter Ketch]

But what about the simplest possible set-up: two equally bright sources, in phase, equidistant from the screen? Will there be points on the screen satisfying the conditions now?

Say the distance between the sources is d, and the distance from the center of the two sources to the center of the screen is L. The position on the screen from the center is x. The two distances to a point x on the screen are
s1 = √{L² + (x + d/2)²}
and
s2 = √{L² + (x - d/2)²}

The ratio (s1/s2)² is 1 at the center and approaches 1 at large x. It has a maximum difference from 1 at x = d/2 where it is (1 + (d/L)²))

So your condition from post 8 will not occur anywhere on the screen if the amplitude ratio is greater than this ratio. Most of the time we consider cases where L >> d and this term is never far from 1

Nevertheless, unless the sources are radiating in all directions, the field will vary as a function of angle and I am sure there will be places away from the center of the screen where the amplitudes become equal. However in the central region no dark fringes occur.

It may be an over simplification to make a blanket statement that if the amplitudes are different no dark fringes will occur, but that is a reasonable shorthand for what will be observed in most practical cases

 

[haruspex]

a blanket statement that if the amplitudes are different no dark fringes will occur,

I came to the opposite conclusion.
If the source amplitudes are exactly the same then the only way a completely dark spot can occur if it is equidistant from the sources. But if the sources are in phase that means a bright fringe, not a dark one.
On the other hand, if the source amplitudes are different, even in the slightest, then there are excellent prospects of finding a point that satisfies my criteria. Whether there is guaranteed such a point I have not determined.

 

[Cutter Ketch]

I came to the opposite conclusion.
If the source amplitudes are exactly the same then the only way a completely dark spot can occur if it is equidistant from the sources. But if the sources are in phase that means a bright fringe, not a dark one.
On the other hand, if the source amplitudes are different, even in the slightest, then there are excellent prospects of finding a point that satisfies my criteria. Whether there is guaranteed such a point I have not determined.

I think you must have missed the point that if the intensity ratio is > (1 + (d/L)²) your criterion cannot be met anywhere on the screen. As I also noted in most cases L >> d, so that maximum ratio is nearly 1 so it takes very little intensity difference to assure your criterion is never met.

Regarding your observation that perfectly equal amplitudes only have an opportunity to perfectly cancel in the middle of the screen, I agree. That doesn’t contradict anything discussed thus far. In fact it supports most of this long discussion regarding sources NOT necessarily making totally dark fringes. I wouldn’t want a discussion about this new interesting fact to confuse any of the previous discussion, so I will only mention that in many practical cases including the two source case discussed above the difference from perfect totally destructive interference is very small.

 

[haruspex]

I think you must have missed the point that if the intensity ratio is > (1 + (d/L)2) your criterion cannot be met anywhere on the screen.

Yes, apologies. I was in a hurry and did not read that part.

 

[songoku]

Thank you very much for all the help (haruspex and Cutter Ketch)