Source-Bound Boundary Scans

Diffraction Order Boundary Three-Row Scan

Higher diffraction order eventually asks for a sine marker beyond the endpoint. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

First order fits inside the sine scale

Aperture width stays five metres and wavelength stays two metres. Order 1 needs target 2 m. The shown sine marker gives 2 m, so the row is accepted.

5 m⋅25=2 mmλ=1⋅2 m=2 mA=15\ \text{m}\cdot\frac{2}{5}=2\ \text{m}\quad m\lambda=1\cdot2\ \text{m}=2\ \text{m}\quad A=1
Diffraction order boundary rowThe aperture width, sine marker, order, wavelength, and acceptance bit are checked.sineslitWidth=5 msineFactor=2/5wavelength=2 morder=1 countacceptedBit=1 bit

Second order still fits exactly

Aperture width stays five metres and wavelength stays two metres. Order 2 needs target 4 m. The shown sine marker gives 4 m, so the row is accepted.

5 m⋅45=4 mmλ=2⋅2 m=4 mA=15\ \text{m}\cdot\frac{4}{5}=4\ \text{m}\quad m\lambda=2\cdot2\ \text{m}=4\ \text{m}\quad A=1
Diffraction order boundary rowThe aperture width, sine marker, order, wavelength, and acceptance bit are checked.sineslitWidth=5 msineFactor=4/5wavelength=2 morder=2 countacceptedBit=1 bit

Third order would need more than endpoint sine

Aperture width stays five metres and wavelength stays two metres. Order 3 needs target 6 m. The shown sine marker gives 5 m, so the row is rejected.

5 m⋅1=5 mmλ=3⋅2 m=6 mA=05\ \text{m}\cdot1=5\ \text{m}\quad m\lambda=3\cdot2\ \text{m}=6\ \text{m}\quad A=0
Diffraction order boundary rowThe aperture width, sine marker, order, wavelength, and acceptance bit are checked.sineslitWidth=5 msineFactor=1wavelength=2 morder=3 countacceptedBit=0 bit