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I was studying Young's double slit experiment (YDSE) in which it was derived that the formula for path difference ($\delta x$) is : $$\delta x = \frac{y \cdot d}{D}$$ Where $y$ is the vertical position of the point where interference occurs, $d$ is the separation between the slits and $D$ is the distance between the slits and the screen.

This formula was derived under the condition that $\lambda << d << D$ (note that $\lambda$ is the wavelength of the light used).

Now my doubt is what happens if $d > D$ ? I had tried searching for answers but couldn't find any.

Thanks !

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The normal pattern that you see is shown in the diagram below.

enter image description here

Each slit produces a diffraction pattern (left hand diagram) which overlap and are seen a one diffraction envelope (red dotted line in right hand diagram) because of the slit separation being small produces.
This diffraction envelope modulates the intensity of the two slit interference pattern (red full line in right hand diagram.

The width of the single slit diffraction pattern is controlled by the slit width and the separation of the interference fringes is controlled by the separation of the slits.

If the slit separation is increased then the separation of the interference fringes decreases.

If the slit separation is large compared with the slit width then the diffraction patterns from the two slits can be resolved as shown in the diagram below.

enter image description here

Interference fringes will now be seen where the light from the two slits overlap as shown in the diagram below.
The separation of the fringes is now much smaller as is their intensity.

If the separation of the slits is further increased there will come a time when it will not be possible to observe the interference fringes.

The Rayleigh refractometer - 1 / Rayleigh refractometer -2 in its simple form is a device which uses the interference from two slits with a relatively large separation.

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If d>D, practically none of the light from the two slits will overlap on the screen. Without overlap there is no interference.

If the slits are extremely thin, light will spread over a wider angle from each slit, so D can be smaller and still allow overlap -- and interference -- between the light from the two slits.

To get a good sense of how the slit width affects the beam spread from a slit, see Fraunhofer diffraction.

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  • $\begingroup$ S.McGrew's answer makes sense to me, but how do we arrive at the upper limit slit separation being equal to D? I understand that, say the slit separation is less than the wavelength, then interference will not occur as the wavelets eventually end up merging into a single wave, and also that if the separation becomes too large, it is possible that there will be a point where it will not be possible to observe the fringes( like Farcher explains in his answer ). But I cannot intuitively understand why that the upper limit is D. Where am I thinking wrong? $\endgroup$
    – Sevfeynn
    Commented Nov 5, 2022 at 19:26

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