Young Double slit experiment condition

[Raghav Gupta]

Can @Drakkith , @Doc Al and others help me in this?
In YDSE,
if s is the size of source slit and S is the distance between source slit and the double slits,
Then why condition s/S <= λ/d must be satisfied to observe fringes?
Here λ is wavelength of light source and d is the distance between two double slits.

 

[blue_leaf77]

You can look up about van Cittert-Zernike theorem for propagation of spatial coherence. That inequality condition follows from the requirement that the transverse coherence length at the plane of double slit produced by an incoherent source of size s must be bigger than or equal to the slit separation.

Just for intuitive thought, as you place the double slit closer to the single slit (S is getting smaller), the incoherent property of the source is getting more and more apparent to the double slit. In this case the visibility of interference pattern will degrade.

 

[Raghav Gupta]

I looked up that Cittert- Zernike theorem in wikipedia but it was not showing any formula that I have written in first post.
From where that inequality has came?

 

[blue_leaf77]

What you can use from the mentioned theorem is a relation that governs transverse coherence length of an incoherent source with size s as observed at some distance S. This relation looks like $y_c = \lambda/\theta_s$ where $\theta_s$ is the so-called half illumination angle, an angle subtended by the source size. Now back to the requirement that the transverse coherence length at the double slit plane must be larger than the slit separation, you should be able to see that that inequality should prevail, even though perhaps with a different in prefactor.

 

[Raghav Gupta]

So here transverse coherence length (y_c)= λ/ θ_s
Now I think that half illumination angle is equal to s/S as angle = arc/radius for small angles.
Now you say y_c > d
So
λS/s > d,
Which gives
s/S < λ/d

Is that true?
Does that theorem of Cittert- Zernike tells that y_c > d ?

 

[blue_leaf77]

You don't need the theorem to see that inequality must be fulfilled. The theorem merely derives the expression for transverse coherence length $y_c$. Coherence area is defined as an area within which any pair of points maintain its phase relationship forever, if these points interfere they will produce high visibility interference pattern. This means the two slits must lie within the coherence area in order to see a clear interference pattern.

 

[Raghav Gupta]

Is that correct θ_s = s/S in 4th post, where I have used angle = arc/radius ?
What is the prefactor you are talking in 4th post?
So can you give a link explaining the relation of the theorem telling that y_c = λ/θ_s?

 

[blue_leaf77]

Yes it is close to correct except that $\theta_s = s/2S$ as $\theta_s$ is defined as the half angle. But the presence of two is not strict as you see if you omit this two you will have even safer assurance that the condition is met.

 

[Raghav Gupta]

So is that the prefactor of two you were talking about?
Can you give a link for showing the relation of y _c= λ/θs ?

 

[blue_leaf77]

http://www.physics.utoronto.ca/~dfvj/presentations/OSA04.ppt

 

[Raghav Gupta]

I was not able to find that relation in power slide.
Can you tell where it was?

 

[Raghav Gupta]

I recently found that spatial coherence area by van Cittert-Z--- Theorem is
A_c = S²λ² / π s², and also

A_c = λ₀²/Ω , where λ₀² is mean wavelength and Ω is solid angle.
How then by you,
y_c = λ/θ_s ?

 

[blue_leaf77]

If you read Principles of Optics by Born and Wolf, it's found that the diameter of coherence area is $B\lambda/\theta_s$, where $B$ is some number that I don't remember. I guess it's a matter of convenience that people simply neglect the prefactor. Another possibility would be that $y_c = \lambda/\theta_s$ is derived for the case of one dimensional slit. In order to check you have to go to the math such as the one followed in Born and Wolf's book.

 

[Raghav Gupta]

Thanks for the references ( I will check that later ) and your help.