# I Questions about Interference Maxima

1. May 28, 2017

### JohnnyGui

I’ve been wachting a video here on interference maxima when light passes a grating and 3 questions came up.

1.

The lecturer shows the following picture:

He mentions that each purple line from each slit (distance $d$ apart) that meet at the same point on the right side (for constructive interference), will be longer than the previous purple line by whole integers of $λ$ (small green lines). However, isn’t this only the case when the purple lines are parallel to each other? You’ll notice that the further away the purple lines are, the sharper their angles are with the slit line so that the proportionality with nλ will not hold. Perhaps this is all neglectable but in the case of precise measurement, is my reasoning correct?

2.
If all purple lines need to be parallel to each other for them to be longer than the previous one by whole $nλ$’s, how will they meet at the right side to make a constructive interference in the first place? They’re parallel to each other.

3.
Since parallel lines don’t meet at the right side, if one truly wants to seek an angle of any purple line that is a distance of $nλ$ longer than the previous ones, while meeting at the right side in 1 point for a constructive interference, shoudn’t one use the very first purple line as a radius of a circle instead of drawing a triangle and solving for $sin^{-1}(λ/d)$? Like this:

Where the green lines are longer by nλ with respect to each other? In this case one would have to find an equation that shows a relation between the length of the green lines and the angle from the circle center w.r.t. the angle of the first purple line. Nevertheless, for very precise measurement, is this reasoning correct?

2. May 28, 2017

### BvU

Correct. The usual derivation is for $\theta << 1$ .

3. May 28, 2017

### sophiecentaur

The assumption is OK as long as the angles are all small enough. The lines that are chosen for the diagram will not actually be the paths that are followed from slots to any particular point on the screen. The light spreads out in all directions from each slot - particularly for very narrow slots, which have individual diffraction patterns which are pretty much equal amplitude in all of the forward facing sector. The overall pattern is the product of the slit pattern and the array pattern and wide slits will impose a variation of amplitudes of the fringe maxima. See this link for an idea of what I am getting at.
It is easier to identify the nulls than to identify the maxima in such a pattern,. For the nulls, the path differences are an odd number of half wavelengths.
The interference pattern for a number of slits greater than two will only have sharp nulls in the 'near field'. where the paths cannot be considered to be parallel and your 'circle' diagram shows it well. The nulls will be 'filled in' because the path differences at points on the screen are never quite right. Twi slits will give good nulls even in the near field.

4. May 28, 2017

### JohnnyGui

Thanks for the explanation. So I take it my circle diagram shows correctly/more precisely how much each path increases in length (green lines) the further it's away from the given maxima?

5. May 28, 2017

### pixel

Usually this is shown with the purple lines parallel to each other, which, as you assumed, is the correct situation. Then a lens is placed to the right of the slits and the lines converge on a screen a focal length away.

6. May 29, 2017

### sophiecentaur

You don't need a converging lens to produce fringes. There is light arriving at each point on the screen from each slit and you do not 'need' those rays on the diagram to be the ones that do the interfering.
I still say that it is better to be examining the minima because they are better defined. It's only a matter of choosing odd multiples of λ/2 instead of even values. No one uses maxima when good minima are available.

7. May 29, 2017

### JohnnyGui

I understand. But what I merely wanted to know is if my reasoning is correct of how the length increases the further away a line is. Would my radius method be more accurate to get the angle compared to when using the sine method of triangles?

8. May 29, 2017

### sophiecentaur

I am not sure what your "radius method" is for - except to exaggerate the path differences for the purpose of a qualitative demonstration. The angles don't count here; it's the path distances between slits and screen point that you use to find the sum of the contributions at any point. It's a Pythagoras job to work out the paths to get the phases. Introducing your radii just means more calculations.
This sort of thing is easy on a spreadsheet and you can get a graph of the resultant amplitudes. Then you can see how the pattern goes from far field to near field as you vary the throw of the rays.

9. May 29, 2017

### pixel

I assume the OP is asking about Fraunhofer, or far field, diffraction, which involves parallel rays. A lens selects rays that are parallel and therefore can produce the expected diffraction pattern on a nearby screen. It's true that if the path length difference between light from the first and last slit is << λ, then it's a good approximation that the individual rays are parallel. Otherwise, and for best far field results, a lens is useful.

10. May 29, 2017

### JohnnyGui

Exactly, so using the radius approach to calculate the increase in lengths of the paths as they're further away should yield the same result as when using pythagorean theorem, albeit in a more complex way? All I want to know is if this radius approach shows the correct increase in path lengths as any other precise measurement.
Doesn't converging the rays after they pass the slits still result in different path lengths such that the sin formula would give less accurate results compared to when the rays stay parallel?

11. May 29, 2017

### pixel

The rays that converge a focal length away from the lens are the ones that were parallel entering the lens. The lens is in effect selecting those rays.

12. May 29, 2017

### sophiecentaur

How would you envisage calculating the lengths outside the circle? I couldn't think of a way that wouldn't involve knowing the total distance to start with.

And what would that achieve? I can't see that there would be any difference in essence. Do you think it would make the nulls sharper with the lens? How would you be able to prove that? The distances from the slits to a point of the screen would be different with the lens there but would the nulls be any sharper? You would need to come up with a geometrical proof of an improvement.
I'm just suggesting that the least complicated method would be the best to go for.

13. May 29, 2017

### JohnnyGui

Indeed, just noticed that you'd need the initial distance of the circle radius to begin with which is dependent on where the next maxima is that you're trying to calculate in the first place.

14. May 29, 2017

### pixel

OP is using the equation for Fraunhofer diffraction. That approximation is for parallel rays exiting the slits. Without a lens, a nearby screen would show a different diffraction pattern based on different equations (Fresnel diffraction). To get a proper Fraunhofer pattern, the screen either has to be very far away or you can use a lens and a nearby screen. So what the lens achieves is that you get the expected Fraunhofer diffraction pattern with maxima at d sinθ = nλ.

15. May 29, 2017

### JohnnyGui

Ah, got you now. However doesn't each slit produce several rays in all directions/different angles? Such that, you can't make all of them parallel with a lens? Just one set of rays, 1 from each slit?

16. May 29, 2017

### pixel

Yes, rays exit the slits in all directions. But at a location on a screen an infinite distance away, what's relevant is the subset of rays leaving the slits in that direction - these would be parallel to each other as the screen is an infinite distance away. At another location on the screen, another subset of parallel rays is involved.

You're not making rays parallel by using a lens, you using the lens to make each of these subsets of parallel rays converge to different locations on a nearby screen.

17. May 29, 2017

With a converging lens, you put the screen in the focal plane of the lens. Parallel rays come to focus in the focal plane at a location $x$ that depends upon the incident angle of the parallel rays. In the small angle approximation $x=f \theta=f sin(\theta)$ where $\theta$ is measured in radians. $\\$ Notice also the lensmakers equation $\frac{1}{f}=\frac{1}{b}+\frac{1}{m}$. When $b=+\infty$, which is indicative of parallel rays, the image distance $m=f$. (In this case we are focusing a diffraction pattern so the source (the set of diffraction slits) is not at $b =+\infty$, but we are interested in where parallel rays focus, which is why we chose $b=+\infty$.)

18. May 29, 2017

### sophiecentaur

A single thin slit produces a very wide (semicircular) beam. There is light coming out in all directions. How could it only come out in 'certain' directions and form fringes by interfering with other similar wide beams from other slits?

19. May 29, 2017

### sophiecentaur

Right. And does that produce sharper fringes? I can see what you say. I also see, now, how it works in the near field!! Right!!!!!

20. May 29, 2017

And the focused pattern is brighter because in the far field, on a faraway screen, the light reaching the screen is often very dim because the screen in the far field is so far away. The irradiance (watts/m^2), falls off as an inverse square.

21. May 29, 2017

### JohnnyGui

Ah of course, for some reason in misread that it makes the rays parallel.

22. May 29, 2017

### JohnnyGui

I was only talking about the rays that were about to make a constructive interference at a particular maxima. If you "break" the semicircular beam to another direction with the lens, another set of rays that weren't about to make a maxima would now do.
This aside, I now see that pixel meant a converging lens.

23. May 30, 2017

### sophiecentaur

It's important to remember that there are no 'rays' as such. They are only aids to calculate what happens. The reason that the lens helps is that it produces plane wavefronts at its focus so that the contributions from each slit have equal steps in phase shift whereas the steps are not equal without a lens (only at infinity).
Thanks to @Charles Link for making me revisit this in my mind and understand it again (50 years later!!!)