Young's Double slit experiment with different slit sizes

[PumpkinCougar95]

What would happen if a youngs double slit experiment is done with different slit sizes? I get the feeling that the intensity might never be zero as the waves might never cancel out completely. Am I right? How should I analyse the Intensity?

Would I have to do something like what is done while analysing the single slit diffraction pattern?

 

[mfb]

Would I have to do something like what is done while analysing the single slit diffraction pattern?

Right. You get single-slit and double-slit effects at the same time. The total interference pattern can be expressed as product of the two patterns.

 

[PumpkinCougar95]

So basically i would get something like this ?
$$ Amplitude = K_1 Sin(Cx) + K_1 \frac {Sin(Bx)} {Bx} $$

where $K_1,K_2,B,C$ are some constants
If I do get this, is there no simple to just calculate the min and max amplitude and where they will occur?
(I don't want to do a lot of computation)

 

[mfb]

Product, not the sum.

If the slit separation is large compared to the slit width (it typically is), then the maxima will be very close to the maxima of the double-slit pattern.
Here is an example

 

[PumpkinCougar95]

Product, not the sum

I am sorry, But why would we take the product of two amplitudes to get the resultant amplitude?

 

[mfb]

$\displaystyle A(x) = K_1 \cos(Cx) \cdot \frac {\sin(Bx)} {Bx}$

Multiply, not add. And use a cosine for the double-slit part to get the maximum at the right place. Example graph

 

[PumpkinCougar95]

Yeah, the graph looks good. But Why? Why are we taking the product and not adding like we normally do with interfering waves?

 

[jtbell]

Why [...] not adding like we normally do with interfering waves?

Because the "single slit" and "double slit" are not separate sources. The slit width and slit separation are two aspects of the same set of sources.

Do you know how the single-slit (with finite width) intensity pattern is derived? If not, I suggest you look that up. The derivation for the double slit (with finite width) is an extension of this. Both of them, in effect, add an infinite number of waves, not just one or two.

Tip: both of these patterns assume something called the "Fraunhofer approximation", so it will probably help to include the word "Fraunhofer" in your search phrases.

 

[tech99]

What would happen if a youngs double slit experiment is done with different slit sizes? I get the feeling that the intensity might never be zero as the waves might never cancel out completely. Am I right? How should I analyse the Intensity?
Would I have to do something like what is done while analysing the single slit diffraction pattern?

It appears to me that whatever the width of the slits, provided they are identical, they will each have a phase centre, and when the path lengths from each phase centre to a point on the screen differ by 180 degrees, there will always be a null.

 

[PumpkinCougar95]

But the slits are not identical.

 

[mfb]

Why not? Usually double-slit experiments are done with two slits of identical width.

 

[PumpkinCougar95]

Yeah but that is the question...What would happen if the slits have different sizes and would the intensity ever be zero?

 

[mfb]

Ah, I thought you would compare multiple different setups where the two identical slits have different widths between them.

You can calculate the integral over all phases to get a formula. Qualitatively: One slit will contribute a larger amplitude, you don't get perfect minima from the double-slit pattern any more. You also get regions where one slit does not contribute (from its single-slit pattern) but the other does, leading to a constant amplitude. In general everything gets messy.

 

[jtbell]

I originally interpreted your question the same way mfb did.

In the general case you'd probably have to do the integral numerically. It would be an interesting exercise to set up a parametrized graph where you can vary the width of one of the slits and see what happens. Any takers?

 

[Andy Resnick]

Yeah but that is the question...What would happen if the slits have different sizes and would the intensity ever be zero?

It's not that hard to figure out, if the incident wavefront is a plane wave. For two slits, widths 'a' and 'b', located at x-x_a and x + x_b is:

E = |a|e^-2πiaξ sinc(aξ) + |b| e^+2πibξ sinc(bξ), where ξ is the conjugate variable to x. This reduces to the 'usual' diffraction pattern when a = b. The intensity is simply |E|^2, which does have an interference term, but it's not obvious if there is always a zero because of the two different sinc^2 terms.

 

[PumpkinCougar95]

Have you used quantum mechanics or something to arrive at this result (because I see iota's in your expression)? because I don't know much about QM. And what is a conjugate variable?

You can calculate the integral over all phases to get a formula. Qualitatively: One slit will contribute a larger amplitude, you don't get perfect minima from the double-slit pattern anymore. You also get regions where one slit does not contribute (from its single-slit pattern) but the other does, leading to a constant amplitude. In general, everything gets messy.

so is there any simple way to just calculate the minimum and maximum in intensity then? Also, why would the amplitude of the waves be different?(since its the wave with the same intensity so shouldn't amplitude be the same too? and just powers are different)

 

[Andy Resnick]

Have you used quantum mechanics or something to arrive at this result (because I see iota's in your expression)? because I don't know much about QM. And what is a conjugate variable?

Nothing so fancy- a basic result from scalar diffraction is that the fair-field pattern is the Fourier Transform of the aperture. A Slit, that is a 1-D opening of width 'a' centered at position 'x0', has a transmission function commonly written as T = Rect ((x-x₀)/a)

https://en.wikipedia.org/wiki/Rectangular_function

By 'conjugate variable', that's the Fourier-transform pair. For example, temporal frequencies pair 't' and 'f'. Spatial frequencies work the same way, but the conjugate variable is an angle.

 

[PumpkinCougar95]

I am still confused. did you take the Fourier transform of something like $$T = Rect(\frac{x-x_a}{a}) + Rect(\frac{x-x_b}{b})$$ how did you get that result ? (Sorry if i am being stupid i only learned about Fourier transform a few days ago)

 

[Andy Resnick]

I am still confused. did you take the Fourier transform of something like $$T = Rect(\frac{x-x_a}{a}) + Rect(\frac{x-x_b}{b})$$ how did you get that result ? (Sorry if i am being stupid i only learned about Fourier transform a few days ago)

Yep, you got it- almost, check the sign, since one is on the other side of the origin (although in the end, it's just a different phase factor)

$$T = Rect(\frac{x-x_a}{a}) + Rect(\frac{x+x_b}{b})$$

 

[tech99]

Right. You get single-slit and double-slit effects at the same time. The total interference pattern can be expressed as product of the two patterns.

Qualitatively speaking, we see a fine pattern resulting from the slit spacing and an envelope resulting from the slit width. This envelope will have a main lobe and small side lobes.
If the slits differ in width, then the fine pattern will have minima which are partly filled in. The envelope from each slit is now different in width, the narrower slit having an envelope which looks similar but is wider. The two overlapping envelopes will now form a "slow" beat pattern, which will probably not be fully visible.
The maximum possible field strength in the pattern will be the sum of the two field strengths measured at the screen.
Notice that a wide slit will tend to generate a parallel beam, which might not overlap the beam from the other slit. This is because the screen is now in the Radiation Near Zone of the aperture.

 

[Charles Link]

The interference pattern from $N$ equally spaced (narrow) slits has the formula for intensity $I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)}$, with $\phi=\frac{2 \pi d \sin(\theta)}{\lambda}$.$\\$ When the denominator is zero, the expression is taken as the limit of where it approaches zero. In those cases, they are the primary maxima, with $m \lambda =d \sin(\theta)$, for integer $m$,and the intensity is $I=N^2 I_o$. $\\$ This result gets multiplied by the diffraction pattern for a single slit of width $b$ when the slits have finite width: $I_D(\theta)=\frac{\sin^2(\frac{\pi b \sin(\theta)}{ \lambda})}{(\frac{\pi b sin(\theta)}{\lambda})^2}$.(Note the form of this expression: $I_D(x)= \frac{sin^2x}{x^2} \,$ ). In the limit of a very narrow slit $b$, the diffraction factor is 1. Otherwise the interference pattern from $N$ slits is reduced by the diffraction factor, which has zeros at $m\lambda = b \sin(\theta)$ for non -zero integer $m$. When $m=0$,$(\theta=0 )$, this is the central maximum of the diffraction pattern, and the diffraction factor is equal to 1. $\\$ And these results are not of quantum mechanical origin. They are classical Kirchhoff-Fresnel diffraction theory. $\\$ Editing: It should be mentioned that these intensity results are for the far-field=far from the slits, and the name Fraunhofer is often associated with these far-field results.