Interference Intensity: Questions & Answers

[BillKet]

Hello! I am a bit confused by the formula for light intensity in the case of interference. In the books and online resources that I read, this is given as: $$I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda})$$ where $d$ is the distance between the slits, $\lambda$ is the wavelength of the light and $\theta$ is the angle where we do the measurements. But in this form it looks like the intensity will be equal to the maximum value $I_0$ an infinite number of times. Shouldn't the intensity maximum go down with the distance from the center, as the energy stored in the wave gets diluted as $1/r^2$ so the further you are the less energy you have at a maximum, compared to the center? Also, as the distance between 2 maxima is equal, it looks like you have an infinite number of maxima, each with the same intensity, which means that the total energy associated with these maxima is infinite, for any amount of initial energy, which doesn't make sense. Can someone explain this to me? Thank you!

 

[Charles Link]

This is for the case of two sources separated by a distance $d$, and it is a formula that applies to the intensity observed in a horizontal plane at angle $\theta$. If you do a little mathematics, you can see that peaks occur for $m \lambda =d \sin{\theta}$, where $m$ is any integer. If $d >> \lambda$ , you get many, many closely spaced peaks, but this becomes an impractical case for interference. $\\$ Energy is conserved for these interference patterns, but to show that, it is necessary to compute the pattern over the solid angle, (e.g. in spherical coordinates $I=I(\theta, \phi)$). It is possible to show that the total power radiated over the $4 \pi$ steradians from two sources that make an interference pattern is equal to the sum of the individual powers.

 

[BillKet]

This is for the case of two sources separated by a distance $d$, and it is a formula that applies to the intensity observed in a horizontal plane at angle $\theta$. If you do a little mathematics, you can see that peaks occur for $m \lambda =d \sin{\theta}$, where $m$ is any integer. If $d >> \lambda$ , you get many, many closely spaced peaks, but this becomes an impractical case for interference. $\\$ Energy is conserved for these interference patterns, but to show that, it is necessary to compute the pattern over the solid angle, (e.g. in spherical coordinates $I=I(\theta, \phi)$). It is possible to show that the total power radiated over the $4 \pi$ steradians from two sources that make an interference pattern is equal to the sum of the individual powers.

Thank you for your reply! So I am aware that the energy is conserved, I am just not sure where does the formula for intensity breaks down. If you measure, say, the intensity of the main peak, you get a finite value. Now, the formula for where the other peaks will be imply that for an infinitely long flat screen, you would have infinitely many bands of intensity equal to the one at the center. So infinitely many bands with finite intensity each would give an infinite intensity. Where is the logic breaking down?

 

[Charles Link]

$\theta$ basically goes from $-\frac{\pi}{2}$ to $+\frac{\pi}{2}$. For these intensity formulas, the total power $P \approx \int I \, d \theta$. To keep the same distance $r$, you need a spherical screen. $\\$ If you collect the energy with a flat screen, you have an inverse square law, and also the screen on the periphery receives the energy at an angle, so that the projected area (perpendicular to the sources) is the relevant area that must be used in the computations. In mathematical terms, for a flat screen, total power $P=\int \frac{I}{r^2} \,dA_{projected}$. In this integral, $dA_{projected}<dA$. The power $P$ that gets computed is completely consistent, and there is no problem with any infinities.
When you get rapid oscillations in the intensity for $d >> \lambda$, essentially the intensity $I$ becomes the average value of $I_o \cos^2$ which is $\frac{I_o}{2 }$.

 

[sophiecentaur]

But in this form it looks like the intensity will be equal to the maximum value I0I0I_0 an infinite number of times.

Where does your "infinity" come from? I think that could be the root of your confusion. Slits, by definition, are all directional. For simplicity, think in terms of two omnidirectional (in XY plane) ideal radio antennae and the maxima will all have the same amplitude. The trig functions in the formula all repeat every 2π so there will only be a finite number of maxima.
You can relax; Energy Conservaton rules OK.
Edit - do the sums yourself for two point sources, separated by only a wavelength or two and look at the pattern, particularly around the direction of the line of the sources; you get mirror image patterns, front to back with sometimes a broad maximum and sometimes a minimum / zero out to the side.

 

[jtbell]

Shouldn't the intensity maximum go down with the distance from the center, as the energy stored in the wave gets diluted as 1/r21/r^2 so the further you are the less energy you have at a maximum, compared to the center?

Yes, if you put a flat viewing-screen in front of the slits, points on the screen that are further from its center are also further from the slits, so you should expect a reduced intensity. The linear width of the interference peaks on the screen becomes wider because of the geometry of the setup.

Another factor comes into play here. The formula you gave is for an idealized two-slit setup in which the slits are infinitesimally narrow, effectively pointlike. Real slits have a finite width. Start with the single-slit intensity distribution:

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html

With two slits of equal width, this distribution is basically "modulated" by your two-slit formula.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/dslit.html

As the slits become narrower, keeping the same spacing between them, the central maximum of the single-slit distribution spreads out, so more two-slit maxima fit inside it. The intensities of the two-slit maxima decrease as you away from the center, until you reach the first minimum of the single-slit diffraction distribution.

You should be able to find a detailed mathematical treatment in an intermediate-level undergraduate optics textbook such as Hecht or Pedrotti. A Google search for "double slit fraunhofer diffraction" turns up some lecture notes which look promising.

Now, the formula for where the other peaks will be imply that for an infinitely long flat screen, you would have infinitely many bands of intensity equal to the one at the center. So infinitely many bands with finite intensity each would give an infinite intensity.

According to the single-slit factor, the "bands of intensity" (what I call two-slit maxima) do not have the same intensity, but decrease according to the single-slit intensity distribution.

I don't remember if I've ever integrated the entire single-slit distribution from -∞ to +∞. I'd be surprised if the result isn't finite.

 

[sophiecentaur]

With two slits of equal width, this distribution is basically "modulated" by your two-slit formula.

It's multiplicative in the ideal case. The element pattern and the array pattern can be multiplied as long as the elements are all identical and pointed in the same direction. In reality, the distribution over each slit may be a bit different etc. etc.

There are layers upon layers of complexity which can be investigated but the ideal case is the one to start with. That's quite hard enough, first time round. And, first you should work in terms of angles before the pattern is projected on a flat screen.

 

[Andy Resnick]

Hello! I am a bit confused by the formula for light intensity in the case of interference. In the books and online resources that I read, this is given as: $$I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda})$$ where $d$ is the distance between the slits, $\lambda$ is the wavelength of the light and $\theta$ is the angle where we do the measurements. But in this form it looks like the intensity will be equal to the maximum value $I_0$ an infinite number of times. [...]

The formula you have above is incomplete, and valid only as a limiting case of each slit width 'a' = 0; the correct formula is:

$$I = I_0 \cos^2(\frac{\pi d \sin \theta}{\lambda}) sinc^2(\frac{\pi a \sin \theta}{\lambda})$$

Where sinc (x) = sin(x)/x.

The sinc term acts as an 'envelope' of the cos^2 function, reproducing the normal decrease of intensity as you get further away from the optical axis.