wider-slit-needs-a-smaller-sine-marker - Wave Optics: Interference and Diffraction Step by Step

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.

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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}