wider-slit-needs-a-smaller-sine-marker

Because the minimum condition a * sin(theta) = m * lambda pairs a and sin(theta) at fixed wavelength and order, widening the slit shrinks the sine marker and pulls the dark fringes inward. The inverse trend is exact only for the stated wavelength, slit width, and order in the ideal single-slit model. Real optics add finite bandwidth, aperture imperfections, alignment error, detector calibration, and measurement uncertainty.

The same wavelength minimum moves to a smaller rational sine marker for a wider slit. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Use a wider slit with the same wavelength

The wavelength and order stay fixed. The slit is wider, so the accepted sine marker must be smaller.

a=10\ \text{m}\qquad \lambda=2\ \text{m}

Wider slits pair with smaller sine markers

Each row keeps the product equal to one wavelength. Doubling the width halves the sine marker.

\begin{array}{c|c|c|c}s&\sin\theta&m&\text{accepted}\\2\ \text{m}&1&1&1\\5\ \text{m}&\frac{2}{5}&1&1\\10\ \text{m}&\frac{1}{5}&1&1\\\end{array}

A wider slit needs a smaller sine for the same minimum

Keeping the wavelength and order fixed, the accepted sine marker gets smaller when the checked slit width is larger.

a\sin\theta=10\ \text{m}\cdot\frac{1}{5}=1\lambda=2\ \text{m}