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Electron double slit diffraction with unequal widths |
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| Apr15-09, 04:25 AM | #1 |
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Electron double slit diffraction with unequal widths
1. The problem statement, all variables and given/known data
Hey guys, I'm working on a problem with a double slit electron diffraction experiment. There is a beam of electrons shooting through two slits onto a screen. When only slit 1 is open, the number of electrons hitting the screen is 25 times the number of electrons hitting the screen when only slit 2 is open. When both screens are open, there is an interference pattern in which destructive interference is not complete. With both slits open, what is the ratio of the probability of an electron arriving at an interference maximum to the probability of an electron arriving at an interference minimum? 2. Relevant equations Unsure. 3. The attempt at a solution This is the first double-slit problem that I've seen, so I have absolutely no idea what I need to do to solve this problem. My instinct would be to just say that since about 25 electrons through slit 1 for every 1 through slit 2, the ratio will just be 26/24 = 13/12, but I'm 99% sure I'm wrong. Assuming this is wrong, why is it? Could somebody please help me out? |
| Apr15-09, 06:25 AM | #2 |
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Recognitions:
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Number of electrons hitting the screen can be taken a the intensity I of the electron beam. Corresponding amplitude of the De Broglie wave is proportional to sqrt I.
Now find I1/I2. Then find A1/A2. During interference Amax = A1 + A2 and Amin = A1 - A2. Hence find Imax/Imin. |
| Apr15-09, 10:14 AM | #3 |
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I1/I2 = 25
A1/A2 = 5 Amax = A1 + A2 = 6, Amin = A1 - A2 = 4 Imax/Imin = (6/4)^2 = 9/4 Is this correct? |
| Apr15-09, 10:28 AM | #4 |
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Recognitions:
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Electron double slit diffraction with unequal widths
Yes. That is correct.
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