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I am familiar with the mathematical aspects of single slit diffraction pattern, at the undergraduate level.

Consider the following pictorial representation from the book Optics, by Hecht:

enter image description here

The fact that I find puzzling here is - even though the slit is shown vertical, the pattern on the screen is shown horizontal. Is this correct? Why so?

My logic:-

The reason why I find this strange is because of a translational symmetry argument. Any two points vertically separated by some distance have the same horizontal attributes. So, one expects the pattern also to have this sort of vertical symmetry, irrespective of what happens along the horizontal axis.

Am I mistaken? If yes, can someone please point out why is the vertical slit producing a horizontal pattern here?

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  • $\begingroup$ Hint: Imagine for a second that the vertical slit is infinite in length together with the focusing mechanism before the screen, how would you then expect the diffraction pattern look and be disposed? $\endgroup$
    – hyportnex
    Apr 7, 2017 at 15:08
  • $\begingroup$ @hyportnex Thanks, but I am very fuzzy about this. Consider any one point of the slit as a single slit. According to Maths, this gives us a sinc function. But I am unable to imagine along which axis? $\endgroup$ Apr 7, 2017 at 15:13
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    $\begingroup$ Think of it this way: How could anything interesting happen vertically if the setup were homogeneous in that dimension? $\endgroup$
    – hyportnex
    Apr 7, 2017 at 15:16
  • $\begingroup$ @hyportnex - I kind of see what you are hitting at. Okay so with one point, we get a sinc function on the z-axis of the diagram, like it is shown. But when you have whole continuum of points in the y-direction above this source point, the whole argument also works for all these points too. So, shouldn't I have central maximum along a whole vertical line, and first minimum along a parallel vertical line, followed by another parallel vertical line for the first secondary minimum etc. (i.e. a pattern having parallel vertical lines, whose intensity degrades horizontally)? $\endgroup$ Apr 7, 2017 at 15:23
  • $\begingroup$ If the illumination is homogeneous vertically, the setup is homogeneous vertically, and the observation is homogeneous vertically then what you will observe must also be homogeneous vertically; so then the only non-homogeneous sinc thing left must be horizontal. So all the lines will be vertical, but now to project that onto the screen you have a focusing lens at $L_2$, that is how you get it on the screen at finite distance. But vertical parallel lines are focused into a single horizontal line by the lens. $\endgroup$
    – hyportnex
    Apr 7, 2017 at 15:29

3 Answers 3

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The wider a slit, the narrower the diffraction pattern. So it makes sense that a tall rectangular slit makes a wide rectangular pattern.

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  • $\begingroup$ Thanks knzhou. This makes perfect sense and goes a long way in teasing an answer, but somehow there is some part of it that's staying unclear to me. Can you please explain the flaw in my reasoning in comment no. 4 below the post? $\endgroup$ Apr 7, 2017 at 15:28
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To observe the diffraction from a slit assume that the vertical dimension of the slit is much larger than its horizontal dimension. To illuminate it you need a line source not a point source otherwise you will get 2-dimensional diffraction pattern not a simple 1-dimensional sinc thing.

The diagram shows that the rays break at $L_1$ and $L_2$, resp., implying a pair focusing (collimating) lenses at $L_1$ and $L_2$ distance. The one at $L_1$ converts the point source into a line source parallel with the long (here vertical) dimension of the slit. The one at $L_2$ collimates the emerging rays from the slit and project them at the screen for observation. The latter lens at $L_2$ focuses the rays that approximately homogeneous in the vertical dimension emerging from the slit into a vertically narrow and mostly horizontally distributed diffraction pattern on the screen for observation. The rays that emerge from the slit are diffracted horizontally but their distribution is nearly homogeneous in the vertical dimension because of the narrowness of the slit.

If there were no lens at $L_2$ you would see a mess on the screen, but if you looked straight into the slit the lens in your eye will collimate and you would see a diffraction pattern but do not do this or you will get badly burned.

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It is the narrowness in the horizontal dimension which cause horizontal diffraction. The slits are only tall because they are not wide.

Incidentally, the diagram is wrong. It shows light entering the full heights of the slits from top to bottom. If it did that, the fringes would be tall too - like vertical lines.

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  • $\begingroup$ Thanks Robert, but let me understand this. When you say horizontal diffraction, you mean the sinc function of the diffraction is along the horizontal axis? $\endgroup$ Apr 7, 2017 at 15:16
  • $\begingroup$ the diagram implicitly assumes a pair focusing lenses at $L_1$ and $L_2$ distance, respectively. $\endgroup$
    – hyportnex
    Apr 7, 2017 at 15:20
  • $\begingroup$ @hyportnex - I get what the first lens is trying to do, but what exactly is the second one doing here? And along which axis is it oriented? $\endgroup$ Apr 7, 2017 at 15:25
  • $\begingroup$ the $L_2$ lens projects the diffraction pattern onto the screen at finite distance from the slit, otherwise you would not be able to see it on the screen because the lines are diverging from the slit. $\endgroup$
    – hyportnex
    Apr 7, 2017 at 15:32
  • $\begingroup$ @UserAnonymous I think a part of the point of this exercise may be that in a camera lens, to shorten the physical length of the lens, one group of lenses near the object excessively converges the light and a 2nd group of lenses near the screen then diverges the light. This means a say 600mm focal length lens can actually be 200mm long say. The aperture is then placed between the lenses. But at very narrow apertures the lens suffers from diffraction issues. This image models that so it's a practical use-case scenario in which diffraction would need to be modelled. $\endgroup$ Apr 7, 2017 at 15:38

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