Three questions about interference pattern and slit.

(i) In the 1-slit experiment, an interference pattern is observed if the slit is wide enough (a narrow slit gives a rather blurred interference pattern). In theory, if we widen the slit more and more but keep the source of light narrow (constant narrow source), the interference pattern will remain sharp or get even sharper. In the limiting case, one can take away the barrier altogether and the slit can become infinitely wide while the source of light is kept narrow (constant narrow source). In this limiting case, the interference pattern will still remain or become the sharpest. Has people done an experiment with no barrier and therefore no slit but with a very narrow source of light to see what interference pattern emerges?

(ii) According to http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide14.pdf, in the 2-slit experiment, the distance of the first minimum (with m=0) from the centre of the central fringe is inversely proportional to the distance between the two slits, i.e., d in eqn. 14.2.9. Using this expression, if we let d decreases towards zero which amounts to the 2-slit experiment becoming a wide 1-slit experiment, the first minimum will be infinitely far away from the central fringe. That would mean that effectively there will be no first minimum and there will be no interference pattern for this 1-slit experiment. But this is contrary to what is observed where we find 1-slit experiment still produces interference pattern. Hence, we have to conclude that there is something wrong with this expression for the position of the first minimum and other minimums. Can someone provide us with a better formula than the one given in eqn. 14.2.9? For that matter, the expression for the maximums in eqn. 14.2.8 also looks suspect.

(iii) According to eqn. 14.2.9, the first minimum and other minimums depend linearly on the distance between the slits and the detecting screen (L). I wonder if people have done experiment by varying L and keep everything else the same, and have observed this linear dependence on L. Since the expression in 14.2.9 seems to be suspect by the argument in (ii), this dependence on L may also be suspect.For that matter, the dependence on L in the expression for the maximums in eqn. 14.2.8 also looks suspect.

  • $\begingroup$ Take a look at Fraunhofer diffraction: [en.wikipedia.org/wiki/Fraunhofer_diffraction] $\endgroup$
    – S. McGrew
    Jan 8, 2019 at 13:28
  • 1
    $\begingroup$ In the 2nd sentence of your 2nd paragraph, you are incorrect in thinking that a two slit experiment can become a single slit. They are completely different and besides it has more to do with the edges than the slits. A single edge produces a fringe pattern and a single slit produces two fringe patterns that overlap to produce a pattern that has destructive interference. A Double Slit produces four single edge patterns that overlap. "See Single Edge Certainty" at billalsept.com $\endgroup$ Dec 10, 2022 at 9:32

1 Answer 1


I think that the formalism of Fourier transforms and Fraunhofer diffraction will be able to answer all your questions. Indeed, when we place ourselves in the case where the distance $L$ between the slit of size $d$ and the screen is very large ($L >> d/\lambda$, where $\lambda$ is the wavelength of the light source), we know that the intensity profile $A(x,y)$ on the screen is related to the Fourier transform $\mathcal{F}$ of the intensity profile $a(x,y)$ at the slit, to be exact $A(x,y)=\mathcal{F}(\frac{kx}{2},\frac{ky}{2})$, where $k=\frac{2 \pi}{\lambda}$ is the wave number. This is the Fraunhofer diffraction.

In the case of the 1-slit experiment, if the slit is infinitely wide in $y$ and of size $d$ in $x$, $a(x,y)$ is a rectangular window function i.e. $a(x,y)=1$ if $x<d/2$ and $a(x,y)=0$ elsewhere. The Fourier transform of a rectangular function of size $d$ is given by $d \frac{\sin(\alpha d)}{\alpha d}$, which give the correct single-slit profile (sine cardinal function). When you take a function and its Fourier transform, a wider function gives a tiner Fourier transform and vice-versa, it explains why it becomes narrow when you increase the size of the slit.

The double slit experiment is quite similar, but the rectangular function is the sum of two rectangular function, if you do the Fourier transform you will observ that the solution is the single slit envelope and some interference pattern. If you decrease the size between the slit, the interference will disapear and the result will be the same as single slit experiment.

enter image description here

  • $\begingroup$ But my objection in (ii) has not been resolved by this answer. $\endgroup$
    – Damon
    Jan 8, 2019 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.