-2
$\begingroup$

enter image description here The book says that the last bright fringe in the interference pattern within the central maxima in the diffraction envelope can not be seen as it cannot be as bright as the diffraction dark fringe.

The diffraction dark spot results due to the destructive superposition of waves giving rise to a dark spot,so any bright spot should be able to overcome this dark spot and the net should be a bright spot.

but in a YDSE,the bright spot cannot be seen as it can not be as bright as diffraction dark fringe i.e the bright spot just disappears,how this happens that a bright and dark spot are added to give a dark spot.

Improve this question
$\endgroup$
6
  • 4
    $\begingroup$ Please include a direct quote from the book to support: "The book says the last bright fringe in the interference pattern with in the central maxima in the diffraction envelope can not be seen as it can not be as bright as the diffraction dark fringe". $\endgroup$
    – S. McGrew
    Commented May 24, 2019 at 18:34
  • 1
    $\begingroup$ "YSDA" is not a standard acronym. What do you mean by it? It is also important that you specify what book you're actually using. $\endgroup$
    – Emilio Pisanty
    Commented May 24, 2019 at 18:40
  • $\begingroup$ Emilio Pisanty YSDA is young's double slit experiment,sorry not to add the same. $\endgroup$
    – sachin
    Commented May 24, 2019 at 19:07
  • $\begingroup$ S. McGrew the book is University physics by sears and zemansky,the book just says the last maxima in the interference pattern cant be seen as it overlaps the diffraction minima,the image attached is from class xiith ncert book of indian curriculum. $\endgroup$
    – sachin
    Commented May 24, 2019 at 19:15
  • 2
    $\begingroup$ What is there to explain beyond what Sears & Zemansky (and Young) say in the book? What don't you understand about their explanation? Or do you simply not believe them? $\endgroup$
    – Bill N
    Commented May 25, 2019 at 1:02

2 Answers 2

Reset to default
1
$\begingroup$

If you have a single slit it will produce a diffraction pattern where there will be maxima and minima in intensity.
So the light from a single slit has "preferred" directions of travel.

Another single slit close to the first one will also produces a diffraction pattern.

If the two single slits are relatively close together the two diffraction patterns from each of the slits will (almost) exactly overlap one another such that there are still regions where the light intensity is a minimum.
The important thing to realise is that presence of one slit does not affect the position of the diffraction pattern of the other slit.

You now have light coming from each of the two two coherent sources (the two slits) which overlap and this produces regions where there the light intensity is a maximum and regions where the light intensity is a minimum - the interference fringes.

In a region where there is no light coming from either of the slits there is no superposition of waves and so no interference.

The image below shows what happens in practice and you can see the intensity of the interference fringes modulated by the diffraction envelope and that the interference maximum at the position of minimum light intensity "missing" because no light is arriving in that region.

enter image description here

Improve this answer
$\endgroup$
9
  • $\begingroup$ thanks a lot,may I know the significance of I(0) i.e intensity in the graph,when the slit becomes very narrow the diffraction pattern becomes very flat i.e almost parallel to x axis,in that case the area under the curve increases,does it mean net energy entering the screen increases,thanks. $\endgroup$
    – sachin
    Commented May 25, 2019 at 10:31
  • $\begingroup$ @sachin I assume that $I_0$ is the intensity of the central fringe. The value of $I_0$ would get smaller as the slit width decreased as less light is getting through the slits and the light is being spread out more. $\endgroup$
    – Farcher
    Commented May 25, 2019 at 12:02
  • $\begingroup$ Farcher 2,I also assume the same but will it change if we plot relative intensity in place of intensity,pl see the picture in physics.stackexchange.com/questions/482089/… ,thanks. $\endgroup$
    – sachin
    Commented May 25, 2019 at 14:16
  • $\begingroup$ @sachin I agree with you. $\endgroup$
    – Farcher
    Commented May 25, 2019 at 14:19
  • $\begingroup$ and one more thing,will we get any pattern if the screen is not exactly opposite to the slits but in the left side or right side i.e if we make the slits in any one wall of a closed room,the other three walls behave like screens,will we get any pattern in the other two walls which are not facing the slits,thanks. $\endgroup$
    – sachin
    Commented May 25, 2019 at 14:21
0
$\begingroup$

A relevant quote from the Khan Academy page you cited is: "If we have a bright spot in the diffraction pattern, then our interference bright spots can be as bright at we want. But, if we have a diffraction dark spot, then the bright spots in our interference pattern cannot be any brighter than the diffraction dark spot, and will disappear altogether."

The authors are referring to "diffraction" as the pattern due simply to a single slit, and "interference" as due to coherent superposition of light from both slits. The quoted passage says, essentially, that if no light gets to a particular point because the point is in a dark fringe of a diffraction pattern from either of the slits, then there is no light from that slit available to interfere with light from the other slit.

That distinction between "diffraction" and "interference" is artificial and misleading. In fact, diffraction is interference and interference is diffraction. They are precisely the same phenomenon observed in vaguely different contexts.

Improve this answer
$\endgroup$
5
  • $\begingroup$ does it mean there is no interference maxima only even though we say that the interference maxima overlaps the diffraction minima,thanks. $\endgroup$
    – sachin
    Commented May 25, 2019 at 14:18
  • $\begingroup$ Yes, in a sense. Actually the answer by @Farcher is excellent. $\endgroup$
    – S. McGrew
    Commented May 25, 2019 at 14:22
  • $\begingroup$ we use long,narrow slits for diffraction,why do we take long slits and is there any limitation to the length of the slits. $\endgroup$
    – sachin
    Commented May 25, 2019 at 14:26
  • $\begingroup$ You should ask that as a new question. $\endgroup$
    – S. McGrew
    Commented May 25, 2019 at 14:46
  • $\begingroup$ question asked separately,thanks,physics.stackexchange.com/questions/482332/…. $\endgroup$
    – sachin
    Commented May 25, 2019 at 16:36

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