### 10.3.1 Explain by means of the principle of superposition, the interference pattern produced by waves from two coherent point sources.

Coherent sources: means that the phase difference between the two sources does not change with time. i.e. if they start in phase they will always be in phase.

When two coherent point sources interfere they produce a interference pattern similar to that shown below:

The dark lines are areas where the two waves interfere destructively, in between the dark points are areas of maximum amplitude where the waves interfere constructively. When the two waves have both traveled an integer value of wavelengths they interfere constructively (assuming the sources are in phase). If the two waves have both traveled an integer value plus one half of a wavelength they interfere destructively.

**Conditions for interference:**

Constructive:$\text{Path Difference} = n \lambda$

Destructive: $\text{Path Difference} = (n + \frac{1}{2}) \lambda$

### 10.3.2 State the conditions necessary to observe interference between two light sources.

### 10.3.3 Outline Young’s double slit experiment for light and draw intensity distribution of the observed fringe pattern.

In 1804 Thomas Young made the first convincing experiment to prove that light diffracted. A sketch of this apparatus is shown below:

For a source he used sunlight, which passed through a filter so it was monochromatic (single colored), the light then passed through a lens designed to create parallel wavefronts (this was required to show diffraction). The light then passed through a single slit and then a double slit. If the light did not diffract the screen at would have been completely dark. Instead of being dark a diffraction pattern was found as shown below:

As a side note: this experiment was very convincing proof that light was a wave, after all how can particles diffract and interfere? They do and it’s a bit weird.

So how do we predict where the light and dark fringes will appear? The geometry of the situation determines where the fringes will appear.

In Young’s experiment the distance between the silts d is very small when compared to the distance between the silts and the screen at point P. Therefore we can treat the lines $\overline{S_1P}$ and $\overline{S_2P}$ as parallel. The line $\overline{S_1Z}$ is drawn such that $\overline{S_1P}$ and $\overline{ZP}$ are identical lengths. This means that the angle $\widehat{S_2S_1ZP}$ is very close to θ. So the path difference of the two sources is $\overline{S_2Z}$. If the path difference is an integer number of wavelengths the sources will constructively interfere and a bright fringe will appear. So the conditions for a bright fringe to appear is:

(1)or re-writing the length $S_2 Z$ in terms of d and $\theta$:

(2)This last equation is given in the IB formula book.

Now lets look at the spacing between fringes. From the sketch below we can state:

(3)Where $y_n$ is the distance from the center point to the nth bright fringe and D is the distance between the screen and the slits.

Because θ is small we can make the approximation $\sin \theta = \tan \theta$ , therefore we can substitute into our earlier equation (2) and get:

(4)Solving for $y_n$:

(5)Now if we look at:

(6)Or in IB notation:

(8)This last formula is in the IB handbook and gives the distance between two successive fringes.

the light sources must be coherent and monochromatic to observe interference between two light sources.

ReplyOptionsAnd what will be the formula if we don't make that approximation, but use exact calculations? These lines are not always "near parallel", and the lines of constructive/destructive inteferences are not straight, but hyperbolas. What then?

ReplyOptionshow can we prove theoretically that the lines of constructive/destructive inteferences are not straight, but hyperbolas??

ReplyOptionsis path difference necessary for the interference ?if there is no path difference then interference will take place or not.

ReplyOptionsYes and no. For (complete) destructive interference a peak (max) of a wave must coincide with a trough (min). This happens most simply due to a path length difference. It could also happen due to a phase difference between two sources. If two sources were at the same point and one source "went up" when the other "went down" they could cancel with no path length difference. Phase difference, not sure if that's covered in IB land, essentially has to do with the timing and coordination of sources or waves, in math terms a phase difference shows up as a horizontal translation of the wave i.e. $y=sin(t)$ vs. $y=sin(t+c)$ where "c" is the phase difference or horizontal shift.

ReplyOptionsNo path difference is necessary. Because the waves spread out there will always be interference, the only thing essential is that the path difference (if there is one) does not change, if it does then the points of cancellation and reinforcement will also change and shift, and if this happens too fast then you won't be able to see any interference pattern.

However if path difference is too great you also won't be able to see interference.

ReplyOptionssir,i want to clear my doubts.these are::

how we get the expressions for constructive and destructive interference,

and

is analytical treatment of interference of light and superposition of waves of same frequency and constant phase difference are the same?

and

describe interference pattern produced due to superposition of coherent sources? reply me

ReplyOptionsThe expression for constructive and destructive interference are generated from a verbal/logical argument.

I am not entirely sure if there is a analytically solution or how to get one. I played around with google and wolfram alpha and didn't get a solution. Solving equations with trig functions often requires either finding one solution (i.e. inverse trig functions) and then generating more solutions with a bit of logic and then testing the solutions OR using computers to search the solution space for an answer.

With that said each point source can be modeled by a function $sin( \sqrt{x^2+y^2} )$. The point sources should be shifted from the origin to see the interference.

You can see the results from wolframalpha http://goo.gl/Yrogf

Your question "is analytical treatment of interference of light and superposition of waves of same frequency and constant phase difference are the same?"

I believe the answer is yes they are the same - if I understand your question correctly.

The description of interference patterns produced by superposition of a coherent source is pretty much covered at the top - if not in words then in picture form.

ReplyOptionsI also cleaned up some of the notation to make the page a bit clearer.

ReplyOptionsnothing is impossible so why two different source of light cannot be coherent , in this world ? and suppose if two different source of light is coherent then what will be the advantages as well as disadvantages ? pl. ans. me!!

ReplyOptionsThe double slit effectively creates two coherent sources of light. I'm not quantum physicist but I think you'd have to work pretty hard to create to separate sources of light that were coherent. So there would be no disadvantage that I can think of other that the extreme unlikeliness of it happening.

-HS

ReplyOptionsWhat the conditions for a light wave to interfere ..? Will it not interfere if it s not coherent and monochromatic? Plz explain

ReplyOptionstwo coherent sources form interference fringe? obtain and expression for the distance between the two constructive bright fringes?

ReplyOptionsHow can obtain two coherent sources form interference fringes? Obtain an expression for the distance between the two constructive bright fringes?

ReplyOptionsNothing is impossible

ReplyOptionsA few things are impossible. Like the comment above being true. :)

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