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Diagram Text

Why does path difference between waves from two secondary sources equal half a wavelength?

And why is $\sin(\theta) = \frac\lambda b$?

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2 Answers 2

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see the second diagram. and there you have the relation.

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The path difference is half a wavelength only fot the first minima fringes, for the next minima we should have path differences of an integer multiple of half a wavelength.

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