Model check. This lesson uses exact small training values (not measured constants) so every relationship is checkable to the digit. Real instruments report measured values with units, significant figures, and uncertainty; the idealized numbers here isolate the rule, not a measurement.

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