integer-path-difference-makes-a-bright-fringe - Wave Optics: Interference and Diffraction Step by Step

A bright interference claim begins with a checked integer path difference. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

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First measure the two path lengths

The two paths are recorded before any fringe label is accepted. Here the longer path exceeds the shorter path by 4 m.

\Delta = 14\ \text{m} - 10\ \text{m} = 4\ \text{m}

Integer wavelength counts stay bright

With the wavelength fixed, each integer wavelength count gives the same bright verdict. The table scans three exact counts instead of relying on one example.

\begin{array}{c|c|c|c}\Delta&\lambda&\text{target}&\text{accepted}\\2\ \text{m}&2\ \text{m}&1&1\\4\ \text{m}&2\ \text{m}&2&1\\6\ \text{m}&2\ \text{m}&3&1\\\end{array}

An integer path difference accepts a bright fringe

The checked path difference is 4 m, exactly 2 wavelengths.

\Delta=4\ \text{m}=2\lambda;\quad \lambda=2\ \text{m};\quad \text{accepted}=1