Physics Help Forum Single-Slit Diffraction

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 May 2nd 2010, 01:54 PM #1 Junior Member   Join Date: May 2010 Posts: 2 Single-Slit Diffraction Hi everyone. I need help comprehending single slit diffraction. I am using the textbook: College Physics by Serway / Vuille 8th Edition. This is what it says: "To analyze the diffraction pattern, it's convenient to divide the slit into halves, as in Figure 24.17 (picture below). All the waves that originate at the slit are in phase. Consider waves 1 and 3, which originate at the bottom and center of the slit, respectively. Wave 1 travels farther than wave 3 by an amount equal to the path difference $\displaystyle \frac{a}{2}\sin\theta$, where a is the width of the slit. Similarly, the path difference between waves 3 and 5 is $\displaystyle \frac{a}{2}\sin\theta$. If this path difference is exactly half of a wavelength, the two waves cancel each other and destructive interference results. Therefore, waves from the upper half of the slit interfere destructively with waves from the lower half of the slit when $\displaystyle \frac{a}{2}\sin\theta = \frac{\lambda}{2}$ The question is: why is a/2 the hypotenuse? why not just use a? I realize the first sentence says, "it's convenient to divide the slit into halves," but why?
 May 30th 2010, 12:22 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 You can .It is just a matter of convenience and symmetry; finding a wave below which cancels a wave above etc. The path difference between 1 and 3 (and 3 and 5) is a/2 sin theta. Thus total path diff between 1 and 5 is a/2 sin theta + a/2 sin theta = a sin theta which is the same as you would get when you use 1 and 5 directly
 May 31st 2010, 03:41 PM #3 Senior Member   Join Date: Mar 2010 Location: Lithuania Posts: 105 I want to share some ideas about the subject. I may be wrong. 1) Consider the slit divided into 3 parts (1, 2, 3). Then the condition of destructive interference for parts 1 and 2 is a/3 sin t = L/2 where L is wave length. Then the sum of parts 1 and 2 at infinity is zero, but part 3 is left. With 3 parts is is hard to find the destructive interference. This model is not good. 2) Consider the slit divided into 4 parts (1-2 1 part, 2-3 2 part, 3-4 3 part, 4-5 4 part (from pic.)). a) Then the condition of destructive interference for parts 1 and 2 is (1) a/4 sin t = L/2. The same for parts 3 and 4. So we may think that at condition (1) at infinity will be zero. b) Consider parts 1 and 3. From condition (1) we get a/2 sin t = L. We can see that this is condition of constructive interference. That is at infinity parts 1, 3 and 2, 4 arrive with the same phase and give there max intensity. So a) and b) give us opposite results. We may conclude that division to 4 parts is not good. We may conclude that two conditions must be fulfilled for finding correct interference: 1) the way of division must be the most simple 2) the result cannot depend on the way of finding the sum. These conditions are fulfilled when the slit is divided into 2 parts.
 Sep 3rd 2010, 11:51 PM #4 Junior Member   Join Date: Nov 2009 Posts: 24 single slit What happens to the distance between fringes as the width of the single slit become larger? Does the distance between fringes increases? What happens to the centre bright fringe? Does it become larger with an increasing single slit width? My hunch is: Increasing the width increases the size of the bright central fringe, but applets on the net show otherwise. Can someone help?

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