Why, for Young’s double slit experiment, does the wavelength have to be greater than or equal to the slit?

Why does the wavelength have to be the same size or greater than the slits when performing Young’s double slit experiment?

<h2>Effect of slit width and wavelength on diffraction patterns</h2>
<p>Using our sketches we see that the extent to which the diffracted wave passing through the slit spreads out depends on the width of the slit and the wavelength of the waves. The narrower the slit, the more diffraction there is and the shorter the wavelength the less diffraction there is. The degree to which diffraction occurs is:\[\text{diffraction}\propto \frac{\lambda}{w}\] where \(\lambda\) is the wavelength of the wave and \(w\) is the width of the slit.</p>
<p>We can do a sanity check on the relationship by considering some special cases, very big and very small values for each of the numerator and denominator to see what sort of behaviour we expect (this is not a calculation, just a check to see what sort outcomes we expect when we change wavelength or slit width):</p>
<ul>
<li>Set \(\lambda=1\) and \(w\) very large, the result will be \(\frac{1}{\text{very big number}}\) which is a very small number. So for a very big slit there is very little diffraction.</li>
<li>Set \(\lambda=1\) and \(w\) very small, the result will be \(\frac{1}{\text{very small number}}\) which is a very big number. So for a very small slit there is large diffraction (this makes sense because eventually you are dealing with a point source which emits circular wavefronts).</li>
<li>Set \(\lambda\) very large and \(w=1\), the result will be \(\frac{\text{very big number}}{1}\) which is a very big number. So for a very big wavelength there is large diffraction.</li>
<li>Set \(\lambda\) very small and \(w=1\), the result will be \(\frac{\text{very small number}}{1}\) which is a very small number. So for a very small wavelength there is little diffraction.</li>
</ul>
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Effect of Slit Width and Wavelength

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Effect of Slit Width and Wavelength

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