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### Course: MCAT > Unit 9

Lesson 16: Light and electromagnetic radiation- Light and electromagnetic radiation questions
- Electromagnetic waves and the electromagnetic spectrum
- Polarization of light
- Diffraction and constructive and destructive interference
- Wave interference
- Young's double slit introduction
- Young's double slit equation
- Young's double slit problem solving
- Diffraction grating
- Single slit interference
- More on single slit interference
- Thin film interference (part 1)
- Thin film interference (part 2)
- Photon energy

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# Diffraction and constructive and destructive interference

Our heros are in a tight spot. One of them smacks the walkie-talkie they are trying to use to contact their lost team member. She laments, “I can’t get a signal! There’s too much interference!”

But what exactly is interference?

## Electromagnetic waves

Both light waves and radio waves are examples of electromagnetic waves, meaning that they fall on the same electromagnetic spectrum as infrared waves (the radiated heat you feel from a stove), ultraviolet waves (the radiation that causes sunburns), and microwaves (the radiation that’s used to cook food in a microwave)

Because these are all waves, they all have a wavelength that determines the distance over which their amplitude changes. Radio waves can have wavelengths as wide as your arms (and even longer!), while visible light waves have wavelengths as small as a thousandth of the width of a human hair

## Wave interference

When two waves come close to one another, their effects add together. If the

**crests**, or highest parts of the waves, line up perfectly, then the crest of the combined wave will be the sum of the heights of the two original crests. Likewise, if the lowest parts of the waves (the**troughs**) line up just right, then the combined trough will be the depth of the two original troughs combined. This is known as**constructive interference**, in which two waves (of the same wavelength) interact in such a way that they are aligned, leading to a new wave that is bigger than the original wave.However, if two waves are not perfectly aligned, then when the crest of one wave comes along, it will be dragged down by the trough of the other wave. The resulting, combined wave will have crests that are shorter than the crests of either original wave, and troughs that are shallower than either of the incoming waves. This is known as

**destructive interference**.In fact, if the two waves (with the same amplitude) are shifted by exactly half a wavelength when they merge together, then the crest of one wave will match up perfectly with the trough of the other wave, and they will cancel each other out. The resulting combined wave will have no crests or troughs at all, and will instead just look like a flat line, or no wave at all!

## Double slit interference

Say you have a laser pointer. A laser is basically just a bunch of light waves that all have the same wavelength and are all lined up with one another. Suppose you place a card in front of the laser beam with two slits in it, such that waves can only pass through two spots. You then measure the amount of light that hits the wall on the other side of the room at various points.

For the experiment to work, the slits have to be tiny compared to the distance from the card to the wall, but they have to be larger than a single wavelength of the light. That means that if we choose a spot on the wall, two light waves will be hitting it; one from the top slit and one from the bottom slit. As they get close to the wall, and close to one another, they will start to interfere. We know that the two waves were exactly the same when they got to the card, but they won’t necessarily be the same when they reach the wall. Let’s choose a spot on the wall to measure the two waves, say above the top slit.

The light coming from the bottom slit has to come much further than the light from the top slit, so more wavelengths will be needed to travel the longer distance. If we choose a different point on the wall, then we’ll get a different number of wavelengths again for each path that the light takes from its slit to the wall. The key is to compare the number of wavelengths it takes for each light wave to travel from the slit to the wall. For constructive interference, the difference in wavelengths will be an integer number of whole wavelengths. For destructive interference it will be an integer number of whole wavelengths plus a half wavelength.

Think of the point exactly between the two slits. The light waves will be traveling the same distance, so they will be traveling the same number of wavelengths. That means that there will always be constructive interference at that spot, so we will always see a bright spot on the wall in the middle.

As you move away from the center point, the two waves’ pathlengths (or total distance travelled from the laser to the wall) will get more and more different, until we hit a point where they are the same plus a half wavelength. At that point, one of the waves will hit the wall with a crest when the other hits with a trough, so they will effectively cancel one another out, resulting in a dark spot there.

As we keep moving away from the center, the pathlengths will keep getting different, until we get to the point where they are the same plus a whole wavelength, so we’ll get constructive interference again, because the two waves will meet at the same spot in their wavelength cycle. This will result in another bright spot on the wall.

This pattern will keep alternating so that we get a pattern of light spots and dark spots, both above and below our center bright spot.

If your slits are further apart, the light waves will be coming from spots that are further apart. That means that their path lengths will be more different from one another, giving bright spots that are closer together.

## Single slit diffraction

You might think that if there is only one slit, there wouldn’t be any wave interference, but let’s say we punched out the whole area between the slits in our card. It’s still a small slit, but it’s much bigger than our slits from the double slit experiment. We can pretend to divide our slit into pieces, and compare the path lengths of the light coming from these pieces to one another to discover what sort of interference pattern we will get when they interact.

Let’s start in the middle of the wall, like we did for the double slit case.

Let’s choose the points at the two edges of the slit. They are an equal distance from the center of the slit, so their path lengths to the center point on the wall will be the same. We know that that means they will interfere constructively with one another.

If we choose two points that are further in, but still the same distance from the middle of the slit, they will also have equal path lengths to the center point on the wall. They will also interfere constructively with one another.

So, we can see that there is a lot of constructive interference going on at that center point, in fact, there will be a major bright spot there because of it.

If we want to find a spot on the wall that is dark, we have to find where there is the most destructive interference. Instead of taking points symmetrically across the slit, let’s take two points, one at the top edge and one just below the center line, and compare them.

Because all of these pairs are the same distance apart across the slit, if we measure the path length from each pair to the same spot on the wall, each pair will have the same difference in path length. (Remember, the pairs won’t have the same absolute path length, just the same difference in path length, which is what we’re interested in anyway.) If we find the point on the wall where one pair has a half wavelength difference in path length, then we’ve found the point where all the pairs will. There will be a lot of destructive interference at that point from all of the different pairs, so we’ll see an overall dark spot there.

Just like for the dark spot, if we find a spot where these pairings have a difference in path length of a full wavelength, we’ll get another bright spot.

If we compare single-slit diffraction to the double-slit interference pattern, the spots are much larger and more spread out. In particular, the center bright spot is much larger than it would be for double slits with the same width. We can view diffraction as light spreading out when it comes up to a hole or other barrier, and trying to get around that barrier. In the process of spreading out, it interferes with itself to create the pattern of light and dark spots that we call a diffraction pattern.

## Double slit interference with diffraction

When we talked about double slit interference, we pretended that only one light wave could go through each slit at a time. If instead we realize that there are a few light waves travelling through each of the two slits at once, then we can see that there will be a diffraction pattern for each individual slit (in addition to the two-slit interference pattern). Since the two slits are close together, and their diffraction patterns are wide, their individual diffraction patterns are similar and we can combine the two diffraction patterns to get the same “single-slit” diffraction pattern that we got for one slit. This pattern will hold our double-slit interference pattern back, limiting how bright the bright spots can be at any given point on the wall.

If we have a bright spot in the diffraction pattern, then our interference bright spots can be as bright at we want. But, if we have a diffraction dark spot, then the bright spots in our interference pattern cannot be any brighter than the diffraction dark spot, and will disappear altogether.

The interference pattern will come from the light from the two slits interacting, and the diffraction pattern will come from the light from each individual slit interacting with itself.

## Consider the following

Imagine our scenario of interference from walkie talkie signals. Say the receiver is between the person sending the walkie talkie signal and a solid rock cliff, and we know that the wavelength of the walkie talkie signal is 1 meter. The signal will reach the receiver, but then keep going past them to the cliff, and bounce off it. The signal will then come back to the receiver. That means that the distance between the receiver and the cliff will determine how the incoming and reflected waves interact with one another. The difference in path length will be the distance past the receiver that the wave travels, plus the distance back again.

Since the wavelength of the walkie talkie signal is 1 meter, then the path difference will need to be a multiple of a whole metre for constructive interference, and a half metre for destructive interference. The receiver can try moving to a place closer or further from the cliff to try to make the signal interference work to their advantage.

## Want to join the conversation?

- For destructive interference, the article says "The resulting, combined wave will have crests that are shorter than the crests of either original wave, and troughs that are shallower than either of the incoming waves" but the diagram shows that the resulting combined wave has higher crests and deeper troughs than either of the incoming waves.(11 votes)
- The representation was not properly drawn for the first example of destructive waves. If anything, it looks more like a partial constructive interference because the waves are slightly out of phase but at most points, they are constructive it seems(2 votes)

- Personally this was very confusing in some parts such as when they were explaining how the lasers would split. How would they split if there's only one laser?(11 votes)
- If there’s only one laser they would be split by light since a laser beam is light and if there was a card with two tiny holes in it then the laser would split and go through both.(0 votes)

- How are electromagnetic waves generated?(4 votes)
- Electromagnetic waves are generated by accelerating charges, also moving charges back and forth will produce oscillating electric and magnetic fields, and these travel at the speed of light.(1 vote)

- For double slit interference, it says that the farther apart the slits, the more different the path lengths and the closer together the bright spots. Why is that the case?(4 votes)
- In Single slit diffraction the article says "If we want to find a spot on the wall that is dark, we have to find where there is the most destructive interference. Instead of taking points symmetrically across the slit, let’s take two points, one at the top edge and one just below the center line, and compare them........" , why can't we continue to take points symmetrically across the slit?won't we get the same result?(3 votes)
- It's basically a mathematical trick, that way you can pair off the all the points. There's a nice video here on single slit interference explaining this.(1 vote)

- what is the equation to find out the wavelength of the light using its interference patter.(2 votes)
- Can someone explain the following paragraph

"in some cases waves do not always cast sharp shadows, such as in Young's double slit experiment the light rays bend out of their straight line path and spreads into the region beyond the slits which would otherwise be shadow. This effect is common for all types of waves, such as sound waves are spread behind the obstacle or corner, which hides the source. Because the sound waves are bending around the obstacle or corner, water waves on a smooth pond or in a ripple tank do not cast clear shadow of an object as the waves are diffracted. "(2 votes) - "For the experiment to work, the slits have to be tiny compared to the distance from the card to the wall, but they have to be larger than a single wavelength of the light"

Why? Can you please explain in detail?(2 votes)- Well, in that case you would still have interference, but the bright regions and dark regions would overlap and you'll end up with a blurry pattern (not very helpful).(1 vote)

- Is there an example of light diffraction that occurs outside of the lab like with late afternoon daylight through textured window pane? Is the photo lower down at this link called 'Remains Of The Day' an example? http://www.graphis.net.au/subartist.php?artistid=58(1 vote)
- a reason why the wavelength of beam cannot be thinner than its diameter(1 vote)