Young's experiment - lowering the light intensity of one slit

[Eitan Levy]

Homework Statement

We change the usual experiment by filtering one of the two slits, so that the intensity of the light that comes out of it becomes half than the intensity of the other slit.
Find the new Imax/Imin in the experiment

Relevant Equations

I is correlated to A^2

I think that what happens is that the amplitude becomes sqrt(2)/2A in the slit filtered, as opposed to A in the other slit.

I suppose we will still get constructive interference from the slits, so the value (1+sqrt(2)/2)A will be reached as opposed to 2A in the usual experiment.

However, the question seems to imply that there will be no dark spots at all. I can't understand why, and how to get the answer.

The final answer is: 33.97

 

[TSny]

I think that what happens is that the amplitude becomes sqrt(2)/2A in the slit filtered, as opposed to A in the other slit.

I suppose we will still get constructive interference from the slits, so the value (1+sqrt(2)/2)A will be reached as opposed to 2A in the usual experiment.

Yes

However, the question seems to imply that there will be no dark spots at all. I can't understand why, and how to get the answer.

When two waves are a half wavelength out of phase, what must be the relationship between their amplitudes in order for them to give complete destructive interference (i.e., to produce zero intensity)?

 

[Eitan Levy]

Yes
When two waves are a half wavelength out of phase, what must be the relationship between their amplitudes in order for them to give complete destructive interference (i.e., to produce zero intensity)?

I understand that in the spots that are dark before we will get (1-sqrt(2)/2)A and that also gives the correct answer. I can't understand why there won't be other points were the intensity will be lower.
Won't there be points where the filtered slit will reach "stronger" than the unfiltered one? There were some points like that before I believe, what happens to them now?
Maybe the question is discussing only the spots where constructive and destructive interference were received?

 

[TSny]

Won't there be points where the filtered slit will reach "stronger" than the unfiltered one?

Yes. There are points where at certain instants of time the filtered wave has a greater value than the unfiltered wave. (Just about all points on the screen have this property.) But there are no points where the intensity I is such that I_max/I is greater than 33.97.

Maybe the question is discussing only the spots where constructive and destructive interference were received?

Can you see that I_min occurs when the waves are $\pi$ radians out of phase? A nice way to see this is with the use of phasors (if you are familiar with this concept).

 

[Eitan Levy]

Can you see that I_min occurs when the waves are $\pi$ radians out of phase? A nice way to see this is with the use of phasors (if you are familiar with this concept).

No, I can't see why this is the case if the waves do not have the same amplitude.

 

[TSny]

No, I can't see why this is the case if the waves do not have the same amplitude.

Consider a point on the screen where the phase difference between the two waves is $\phi$. Let the wave from slit 1 at that point be $A_1\sin( \omega t)$ and the wave from slit 2 be $A_2 \sin (\omega t +\phi)$. Using trig identities or using phasors, you can show that when these are added together, the result is $A_{\rm net} \sin(\omega t +\phi_{\rm net})$, where

$A_{\rm net} = \sqrt {A_1^2+A_2^2+2A_1A_2 \cos\phi}$

$\tan \phi_{\rm net} = \large \frac{A_2 \sin \phi}{A_1 + A_2 \cos \phi}$

The resultant intensity is proportional to $A_{\rm net}^2$. For what value of $\phi$ do you get $I_{\rm min}$?

[Added note: For any point on the screen, there will be instants of time when the sum of the two waves is zero. But, the intensity $I$ at a point on the screen is defined in terms of the time average of the resultant wave.]