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### Course: Physics library>Unit 14

Lesson 2: Interference of electromagnetic waves

# Diffraction grating

What happens when there's way more than two holes? Uncover the power of diffraction gratings. Learn how adding more slits to a double slit experiment results in sharper, brighter interference patterns. Understand why this occurs and its significance in measuring light waves. Created by David SantoPietro.

## Want to join the conversation?

• Why do we have to draw lines at right angle to measure path difference, instead of making an isosceles triangle, in which second side would be equal to length of 1st path? The remaining will be the difference.
• In wave optics we assume that the distance between the slits and the screen is very large as compared to the separation between the slits itself. Hence the angle adjoining the screen in the isosceles triangle about which you are talking will become very small. The other two angles are equal ( a property of isosceles triangles) and each will be approximately equal to 90 degrees since the sum of all the angles in a triangle must be equal to 180 degrees ( and the third angle is very small).
• what is the difference between interference and difraction?
• Interference occors when one wave 'meets' another. (Check out superposition)
Diffraction is the spreading of a wave due to it going through a gap (Aperture) or around a small object. (such as in single slit diffraction)

How do they work together ?
When light passed through the slits in a grating for example, it is diffracted...spreads out towards the screen.
When the light from different slits meet at the screen, the waves will interfere and the resultant amplitudes (determined by superposition) will give pattern on the screen. (known as diffraction pattern)

Hope that helps
• If you make the diffraction gratings closer together, would a line spectrum have higher resolution?
So, if you use λ=dsinΘ, and you decrease d, does that mean sinΘ increases? But it is periodic, so what would happen to the spectrum? Would the spectral lines be closer together?
• The spectral lines would get farther apart, because like you said, "λ=dsinΘ and if you decrease d, sinΘ increases", which means Θ increases. Indeed, by spreading out the spectrum you do achieve higher resolution. So for example, you might imagine there are two spectral lines, and using one grating with a certain d, the two lines are only separated by .1°, so they are too close to be recognized as two separate lines. But then you use a finer grating with d/10 between the grating lines, then the same two spectral lines will be separated by about 1°, and maybe then you can see that there are two distinct lines.
• what is the relationship between diffraction and wavelength of light
• For light (or any kind of wave for that matter) to undergo diffraction, the size of the diffraction grating, or the 'obstacle' faced by the wave should be comparable to its wavelength in size.
This is the reason sound waves diffract around objects(like walls) and you're able to hear the sound but you can't see what's behind the wall because wavelength of light is much smaller compared to the wall so the light gets reflected and not refracted.
• i am getting stuck at the part where David explain the interference at delta x= 1.1*lamda. Can anyone show me a full illustration of how these waves interfere with each other?
• Oh so you're getting stuck on how exactly as we move away from the constructive point do waves get destructive. Here's how: We know from double slit that as you move away from the constructive point the path length difference or ∆x is not exactly 1λ but is slightly more or slightly less. Considering it s slightly more, as in the example David gave in the video, just as in double slit the light coming from one slit was slightly less constructive ( in the example we just set it to be skewed by 1.1 wavelengths) , the light in this case coming from every slit was a little lesser constructive compared to the previous. What I mean is each light wave emerging from the slit was 1.1 wavelengths shifted compared to the previous ( just as in the constructive point you had each light wave being 1 wavelength shifted compared to the other) * and therefore, over a large number of light waves from the many slits, we had each wave cancelling out with the other* . For example the wave that was 1.3 shifted from the first ( and still only 1.1 shifted from the previous) and the one that was 1.8 shifted would destroy each other as the former would be shifted to near a the zero point, i.e. The part of the wave that would reach the point would be near zero if represented in the wavy way, and the latter would have reached the point having being shifted more than π times . Therefore they both destroy each other. And this happens with every wave. Hope this made sense.
• At , you said that the path difference(delta x) will be also be lambda since the angles(theta) are equal. How are they equal?
are they considered to be equal because D>>d. it would be really great if you could tell me using a diagram what angles are you talking about. i figured that path difference between the second and third wave will be lambda using similarities of triangle but am not able to relate the two Thetas youre talking abt.
• They are said to be equal because D>>d as you said , so approximately the lines are parallel, they meet at the same point and hence at the same angle. He explains this in the next video "Single slit interference" at .
(1 vote)
• Can diffraction grating be used for invisibility or cloaking?
• At , If each one is off by 1.1 of a wavelength, wouldn't the number of slits need to be exactly 10 to have complete destructive interference? If in the example instead of 10 slits you had 11 slits, it makes sense that all 10 would create a perfect destructive interference, but wouldn't that last laser through the 11th slit create a wavelength that doesn't get canceled out by anything? Then I don't get why you wouldn't have visible light at that point. Thanks!!
• Is there any real life application for this phenomenon?
• Yes. If you wanted to find the wavelength of the laser you are using use the formula shown at
• But shouldn't this destructive points require there to be infinite holes?
• I believe for a complete destructive interference to occur, the number of points with different phases must range exactly 1 wavelength. Therefore all the amplitudes of the phases will sum to 0. Therefore being fully destructive. Since there are thousands of slits on a diffraction grating, the effect that an extra few phases have on the overall interference in negligible, to an extent that it cannot be seen. So yes, an infinite number of slits would also work.