Derivation of single slit diffraction formula

[songoku]

Homework Statement

This is not homework

Please see below

Relevant Equations

None

I want to ask several questions regarding to the text:
1) Why do we find the minima of the diffraction? Why not the maxima?2) "Figure 25.32b shows two rays that represent the propagation of two wavelets: one from the top edge of the slit and one from exactly halfway down"

Why do we take point exactly halfway down? Why not like a third or a fifth down?3) " The other minima are found in a similar way, by pairing off wavelets separated by a distance of 1/4 a, 1/6 a, 1/8 a, ..., 1/(2m) a"

Again, why don't we take other values like 1/3 a, 1/5 a, 2/3 a?Is the answer to my first and third questions is because that's what we get from doing the experiment? The pattern suits for minima but not for maxima?

Thanks

 

[anuttarasammyak]

1) Why do we find the minima of the diffraction? Why not the maxima?

If you wish you can do it. The obvious maxima is $\theta=0$ , a straight ray, that is boring.

 

[songoku]

If you wish you can do it. The obvious maxima is $\theta=0$ , a straight ray, that is boring.

Why not other maxima besides the center one?

Thanks

 

[anuttarasammyak]

If you wish. I observe in your attached photo the width of the first minimum is narrower that that of maxima. It is convenient to observe. Thanks for the information.

 

[anuttarasammyak]

Why do we take point exactly halfway down? Why not like a third or a fifth down?

In figure (b), the two rays cancel in a pair. Going down from them all the upper rays and downer rays make pairs for cancel so all the rays cancel. It does not work so for starting a third or fifth down.

Again, why don't we take other values like 1/3 a, 1/5 a, 2/3 a?

The chances for cancel follows as 1/4,1/6,1/8

For your favorite maxima why don't you investigate the case m is a half integer 3/2,5/2.7/2... ?

 

[songoku]

I observe in your attached photo the width of the first minimum is narrower that that of maxima. It is convenient to observe. Thanks for the information.

Well, I don't understand why it is convenient to observe, I just post the picture from my book. But you are welcome

In figure (b), the two rays cancel in a pair.

How can we know the two rays from the top edge of the slit and one from exactly halfway down will cancel? From the picture, the text just writes "finding minima". That's the reason I ask why not maxima.

I understand the path difference is $\frac 1 2 a \sin \theta$, and then the text says: "if this extra distance is equal to $\frac 1 2 \lambda$".
I don't understand this part. Why take the "if" as $\frac 1 2 \lambda$, why not $\frac 3 2 \lambda$? And also why we take the path difference to cause minima? Why not equate the path difference to $\lambda$ in the first place?

For your favorite maxima why don't you investigate the case m is a half integer 3/2,5/2.7/2... ?

How to start doing the investigation?

Thanks

 

[anuttarasammyak]

Well, I don't understand why it is convenient to observe, I just post the picture from my book. But you are welcome

Do not take it so seriously. I just thought as attached.

 

[anuttarasammyak]

How can we know the two rays from the top edge of the slit and one from exactly halfway down will cancel?

They cancel because they have wave length difference $\lambda/2$ in path length difference of an end and the middle of slit or phase difference of $\pi$. As you say $\lambda (n+1/2)$ also works for n-th zero point.　
The above sketch is an illustrative explanation for n＝1　\lambda=a \sin\theta
Red line shows E strength, though perpendicular to plane actually, depending on the position. Sum of red vectors = 0. For n>1 there will be n wave trains between which results null in summation.

 

[anuttarasammyak]

How to start doing the investigation?

You know minimum is simple to investigate, it is ZERO.
Maximum is, so to say, the Most Incomplete Cancellation Nearby.
Take "one - thirds case" you mentioned. The first two 1/3 cancel to zero. The remaining one 1/3 survive.
Changing 1/3 to 1/3+delta>0, The cancel portion increases so slit light becomes darker.
Changing 1/3 to 1/3- delta, The cancel portion decrease but new cancel portion appears so in total light becomes darker.
Thus you may know 1/3 case or more in general $\lambda (n+ 1/2)$ in path length for full slit width provides a n-th local maximum. You see for larger n cancel portion increase more so the peak of brightness in maximum going downer.

 

[songoku]

I am really sorry for late reply

I need to re-read all you reply and try to understand it. Thank you very much for the help and explanation anuttarasammyak