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I to be assuming this formula"s derivation involves some degree of approximation, because another formula in the very same section suspect the distance between the slit and also the screen is similar in size to the hypotenuse the picture above.
Using the same approximation, I gained something similar but no the same:$$d = \fracm\lambda\sin\theta \, .$$
Which an outcome is correct??
homework-and-exercises optics double-slit-experiment interference
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edited Aug 3 "15 at 16:15
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If the slits space on peak of each other, then the irradiate travelling through each slit goes the same distance and also therefore has actually the exact same phase.In this case, the distance between the fringes is infinite.On the other hand, if the slits are really far apart, then even a tiny angle incurs a large path difference, so the edge are an extremely close together.Thus we have actually reasoned the the distance in between the edge goes under as $d$ go up.
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Consider the formula written by OP:
$$d = \fracm \lambda\sin \theta \, .$$
The first intensity preferably occurs when $m=1$ giving
$$ \sin \theta_1 = \frac\lambdad \, .$$
Expanding the $\sin$ to lowest order we get
$$\theta_1 = \frac\lambdad$$
which says that increasing $d$ renders the edge of the very first maximum smaller, as we guess above.From this reasoning, we watch that OP"s formula is more than likely correct and also that placing the $\sin$ in the numerator would provide the dorn behavior.
Note that ns didn"t actually derive the correct answer, I just showed that relocating the $\sin$ role from denominator to numerator would more than likely be incorrect.That said, offered the definitions in the question, the exactly formula for the maxima the the two slit interference is in fact