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In my book'a section on Young's double-slit experiment, the formula, $d = m \lambda \sin\theta$, is given. In this equation $d$ is the distance between two slits, $\lambda$ is the wavelength of light coming through the slits, and $\theta$ is the angle between the central reference to the brightest maximum on the screen opposite the slits.

enter image description here

I am assuming this formula's derivation involves some degree of approximation, because another formula in the same section assumes the distance between the slit and the screen is similar in length to the hypotenuse the picture above.

Using the same approximation, I got something similar but not the same: $$d = \frac{m\lambda}{\sin\theta} \, .$$

Which result is correct??

most venerable sir
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  • Sounds like most of this question is actually not relevant to what you want to know. I think the only question is why the angles in the green and red lines are the same. Could you remove the parts that are not needed for that specific question? We try to make questions as clear as possible by removing irrelevant material. – DanielSank Aug 02 '15 at 23:40
  • @DanielSank, I can make another question on mathSE. Asking that. – most venerable sir Aug 02 '15 at 23:42
  • Actually one physicsSE since it has a geometry tag, – most venerable sir Aug 02 '15 at 23:57
  • What is the question? Are you asking which formula is correct? Are you asking how to derive the correct formula? Be specific. – DanielSank Aug 03 '15 at 00:23
  • How to derive? I just gave my own input. – most venerable sir Aug 03 '15 at 00:24
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    You're asking how to derive it, yet you give your own result which suggests that you know how to derive it. I'm not trying to be a pain. I'm trying to help you ask a good question. If you have your own derivation and the question is why the result of that derivation differs from the one in your book, say that explicitly in the question. If the question is about how to do the derivation in the first place, then show us your steps and ask about any parts where you're unsure if your work was correct. – DanielSank Aug 03 '15 at 00:27
  • @DanielSank, I understand you are trying to help. Sorry if I sound rude. – most venerable sir Aug 03 '15 at 00:49
  • Trust me on the edits. I've been using this site for some time and I have a reasonably good understanding of how to write questions effectively. Referring to your book in the title is not a good idea as the reader has no idea (and does not care) what your book says. You want the title to clearly state what is the content of the question. I've even written a self-answered meta post about titles. Also, the period I added after the equation is proper punctuation, so I don't understand why you deleted it. – DanielSank Aug 03 '15 at 00:54
  • But the last one sounded irrelevant as I am trying to find any maximum. – most venerable sir Aug 03 '15 at 00:56
  • Do you still need for information on this? If you think an answer is satisfactory part of the procedure is to mark the check next to it. If you don't think any of the answers are satisfactory, please comment. – DanielSank Aug 04 '15 at 17:16

1 Answers1

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If the slits are on top of each other, then the light travelling through each slit goes the same distance and therefore has the same phase. In this case, the distance between the fringes is infinite. On the other hand, if the slits are very far apart, then even a small angle incurs a large path difference, so the fringes are very close together. Thus we have reasoned that the distance between the fringes goes down as $d$ goes up.

Consider the formula written by OP:

$$d = \frac{m \lambda}{\sin \theta} \, .$$

The first intensity maximum occurs when $m=1$ giving

$$ \sin \theta_1 = \frac{\lambda}{d} \, .$$

Expanding the $\sin$ to lowest order we get

$$\theta_1 = \frac{\lambda}{d}$$

which says that increasing $d$ makes the angle of the first maximum smaller, as we predicted above. From this reasoning, we see that OP's formula is probably correct and that putting the $\sin$ in the numerator would give the wrong behavior.

Note that I didn't actually derive the correct answer, I just showed that moving the $\sin$ function from denominator to numerator would probably be incorrect. That said, given the definitions in the question, the correct formula for the maxima of the two slit interference is in fact

$$d = \frac{m \lambda}{\sin \theta}$$

as written by OP.

DanielSank
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  • So my textbook is wrong! – most venerable sir Aug 03 '15 at 15:17
  • @user132522 looks that way. It would be nice if someone else could confirm this. – DanielSank Aug 03 '15 at 15:55
  • @DanielSank: Okay, sir, could you tell what you meant by even a small angle incurs a large path difference, so the fringes are very close together? Which angle are you talking of? And also, how by a large path difference, the fringe-width becomes smaller? –  Sep 02 '15 at 08:53
  • @user36790 Draw a picture. If the slits are very far apart, then at small $\theta$ (see the picture in the question) the paths from either slit to a point on the screen have very different length. – DanielSank Sep 02 '15 at 14:57
  • Okay, if increasing the distance between two slits make the paths larger, how does it lead to decreasing the fringe-width? –  Sep 02 '15 at 15:04
  • @user36790 Fringes come from the light from either slit going through a different path length. If the path lengths differ by a lot for even small $\theta$, then the fringes are closer together. – DanielSank Sep 02 '15 at 15:05
  • Yes, sir you are absolutely correct but what I want to know is what really happens that decreases the fringe-width when the path-difference becomes large. –  Sep 02 '15 at 15:10
  • @user36790 The fringes happen simply because as you move along the screen, the phase of the light is changing. When the phases add constructively you get high intensity, and when they add destructively you get low intensity. If the path lengths vary rapidly with changing $\theta$ then we run through more phase variation and see more fringes, so each fringe is narrower. – DanielSank Sep 02 '15 at 15:11
  • Thanks, sir; I've drawn two lines to a point when $d \approx 0.5~\text{cm}$- the difference was negligible but when I took $d= 10.25~\text{cm}$, the difference was huge & that's why the interference pattern narrows down as the phase changes quickly then. The opposite thing happened when I increased $D$. Really thanks for the explanation. Lastly, why does the fringe-width increase when I increase the wavelength?? Thanks again:) –  Sep 02 '15 at 15:31
  • @user36790 I think you can figure out the wavelength issue by yourself at this point. – DanielSank Sep 02 '15 at 16:12
  • One question more, sir: Have you noticed the single slit in Young's Double-slit apparatus used for making the source coherent? My book says if it is brought closer to the double-slit board, then the sharpness of the fringes decreases gradually. Can you tell why? –  Sep 02 '15 at 19:11
  • @user36790: That is an entirely new question, and it is a good question. Please post it separately. – DanielSank Sep 02 '15 at 21:35
  • Thanks for the advice, sir; I've posted a new question regarding my query. If you've some time, you may check http://physics.stackexchange.com/questions/203979/why-does-moving-the-source-slit-closer-to-the-double-slit-plane-decrease-the-sha . Thanks:) –  Sep 03 '15 at 05:01
  • @user36790 by the way, a bit of advice: you do not need to call anyone here "sir". There is no notion of authority here. We're all equals. – DanielSank Sep 03 '15 at 05:04
  • @DanielSank: Ha! I never saw that... yeh, we all are equal :) –  May 07 '16 at 16:58