Barrier of Diffraction: Different Wavelengths, Same Color?

In summary: I don't know, 570 or so grooves per mm, we could track the wavelength of each color of light.The problem is that after we did all that measuring and whatnot and had to write a small summary, the Professor said that the wavelengths of some colors (which were in the book, but not in the exercises) don't go in turn like the other colors. instead, they go "bright, dark, bright, dark". And so, I'm having a bit of trouble understanding why it happens.The wavelengths of colors go in turn, but at the Fringe of Confluence, the color we tracked, blue, fades bit by bit. So, it's like we're seeing a "different"
  • #1
Const@ntine
285
18
Hey there! We started Lab III last week, but things are a bit... strange. See, those exercises were written back in the early 1900s,and so they ask us to read from a book that was published around 1890 or so. Naturally, the library has only one copy, and it's open for 2hours per week, so it's not worth the hassle. Anyway, apart from that we have the regular books that the exercises are based on, but they ommit certain stuff,and the Professor of those one or two exercises is one of those "I don't give a damn, that's the book, go search at a thrift shop and find it, but get one question wrong and you're out". So I'm in a bit of a tough spot.

Either way, let's get onto the question itself. In that experiment, we pretty much just studied a Barrier of Refraction with the light-source being Na (Sodium) Steam, and the wavelengths of different colors of certain light-sources, using a Spectrometer. The problem I have is that after taking all the measurements and whatnot, we had to find some stuff through the given equations and write a smalle ssay, and I'm having a bit of trouble understanding why something happens with the former part (BoR).

So, my problem is that using certain equations, and knowing certain data, I have to find the wavelengths of a certain color in two consecutive fringes of confluence. I do that since I know all the data, but my question is: If the color is the same, why do I have a different wavelength in the two different frings? My book doesn't say anything past "at the fringes of confluence it's bright, in the non-confluence parts it's dark, and those go in turns" to give the simplified version.

I could try to post the data and numbers here, but it'd be tough to translate everything, and in the end I think it's just a "theory" question. I thought about it, and the best I could come up with is this:

So, a Barrier of Refraction acts as amix between Difraction and Confluence, right? Taken by themselves, Confluence creates bright and dark spaces on a screen, in turns. A Refraction does the same, but every other bright "spot" fades a bit each time. Put both of them together, and it looks like Confluence, but each "bright light" is now replaced by a series of dark-bright-dark-bright "beams", due to Refraction.

Therefore, at each Fringe of Confluence, the color we tracked, blue, fades bit by bit. And so, it's like we're seeing a "different" color, and thus has a different wavelength. So it's kinda like how Red and Yellow have different wavelengths.

Am I close to anything? Like I said, my book's a bit fuzzy on that (it's a bit difficult for Undergrands I think, it's Serway's 8th Ed. of Physics for Scientists and Engineers), the other book cannot be found and the leaflet with the exercises says even less.

Any help is appreciated!
 
Science news on Phys.org
  • #2
Techno_Knight said:
Either way, let's get onto the question itself. In that experiment, we pretty much just studied a Barrier of Refraction with the light-source being Na (Sodium) Steam, and the wavelengths of different colors of certain light-sources, using a Spectrometer. The problem I have is that after taking all the measurements and whatnot, we had to find some stuff through the given equations and write a smalle ssay, and I'm having a bit of trouble understanding why something happens with the former part (BoR).
Please explain this more precisely. It's not at all clear what you did. If you did an experiment, there must have been a procedure that you followed. What was it? I am sure it didn't instruct you to take "some measurements and whatnot."
 
  • Like
Likes tech99, berkeman and nasu
  • #3
Are you looking at Sodium Vapour or Sodium Flame using a spectrometer?
 
  • #4
kuruman said:
Please explain this more precisely. It's not at all clear what you did. If you did an experiment, there must have been a procedure that you followed. What was it? I am sure it didn't instruct you to take "some measurements and whatnot."
Well, I'll try, but like I said, it's all a bit jumbled up. There were no clear instructions, just "go here ,use this lever and read these numbers".

EDIT: It's a Barrier of DIFFRACTION. I messed up the translation and I can't edit the header now. I don't know why, but I messed up Refraction and Diffraction. Sorry.

So, we used steam of Sodium as a light-source (I don't know how exactly, but that's what they told us), a small piece of glass with 570 grooves per mm and a... box, I guess? Nobody told us what that was. In my book in such instances the writer uses screens (like a tv-screen) so I imagine same went for that box. Anyway, with all of these we created a Barrier of Diffraction. Then, with a spectrometer, we measured the angles (in degrees) of the 0, first and second fringes of confluence of the blue color.

maxresdefault.jpg


slit_array2.png

The experiment looked like the second pic, but I couldn'tfind it in English. Translation from left to right is:

"Laser Source", "Glass Piece with Ridges", "Screen"

We were given certain equations (I'd post them here, but they're just "plug and play" formulas that have to do with a sketch they gaveus-I'd basically need to translate 4-5 pages of a textbook), and with them we were tasked with finding the wavelengths from each fringe of confleunce. My question is, why is there a different wavelength at each fringe of confluence?

tech99 said:
Are you looking at Sodium Vapour or Sodium Flame using a spectrometer?
Sodium Vapour I'd guess, there was no mention of Flame. But the main part is the Barrier of Diffraction that's created.
 
  • Like
Likes mdbablu
  • #5
The source has only one wavelength, and no other wavelengths or colours will be seen.
I think you being asked to measure the angle of each bright fringe and from that to calculate the wavelength of the laser.
Measure the angle between the axis and the fringe. You might have to measure the angle of a group of, say, 10 fringes, and then divide the angle by 10.
 
  • #6
tech99 said:
The source has only one wavelength, and no other wavelengths or colours will be seen.
I think you being asked to measure the angle of each bright fringe and from that to calculate the wavelength of the laser.
Measure the angle between the axis and the fringe. You might have to measure the angle of a group of, say, 10 fringes, and then divide the angle by 10.

Well,I can't go back and measure more angles now. The original exercise had us doing that 5 times (measuring the angles of the first 2 fringes of confluence), but our Professor had us do it only once. As for the rest, I'll trasnlate the question here:

"Light the slit with a light-source and take a look at the fringes of confluence of 1st and 2nd class. Pick a color and measure the wavelength from the fringes of the 1st and 2nd class. The measurement of the wavelengths shall be done by using the formulas 5 & 6."

5: sin(ψ' - φ) = nλ/a - sinφ
6: sin(ψ + φ) = nλ/a + sinφ
(The angles should be "plugged in" in their absolute values)

For more info we have:

ψ' = θ + φ & ψ = θ' - φ

tanφ = (sinψ - sinψ')/(2 - cosψ - cosψ')

a = 18300 A (Angstrom = 10-10m)

1st Fringe:
ψ1 = 20,4 | ψ1' = -8,6 | φ1 = 69.6

2nd Fringe:
ψ2 = 34,7 | ψ2' = -23,6 | φ2 = 32,9

Using either (5) or (6) (I used (6) this time) I find that λ1 = 1150 Α & λ2 = 3490 Α

My question is why does this happen?
 
  • Like
Likes mdbablu
  • #7
Techno_Knight said:
My question is why does this happen?
From the drawing that you posted, it seems that you observed an array of diffraction maxima produced by a two dimensional array of apertures. This happens because of interference. Is that what you are asking about?

Also, the wavelengths that you found are not in the visible range (7000 - 4000 A). You must have made a mistake either in the interpretation of the data or in your calculation of the wavelengths.
 
  • #8
Techno_Knight said:
"Light the slit with a light-source and take a look at the fringes of confluence of 1st and 2nd class. Pick a color and measure the wavelength from the fringes of the 1st and 2nd class.

Why do you have different colors if you're using a laser?
 
  • #9
kuruman said:
From the drawing that you posted, it seems that you observed an array of diffraction maxima produced by a two dimensional array of apertures. This happens because of interference. Is that what you are asking about?

Ah, is that what it's called? I just did a direct translation and went with "Barrier". I get why that happens, the leaflet's question is:

"Light the slit with a light-source and take a look at the fringes of confluence of 1st and 2nd class. Pick a color and measure the wavelength from the fringes of the 1st and 2nd class. The measurement of the wavelengths shall be done by using the formulas 5 & 6."

kuruman said:
Also, the wavelengths that you found are not in the visible range (7000 - 4000 A). You must have made a mistake either in the interpretation of the data or in your calculation of the wavelengths.

Yeah, I noticed that. I'm waiting for my Lab Partner to do his own calculations and check that as well.

------------------------------------
From what I can tell, we have this:

chapter-5-diffraction-34-638.jpg

We looked at the smaller "bump" (I call that a fringe,since I don't know the direct translation), and the one after that (the pic is a random one, just to explain the question). Using the formulas I wrote above and the angles of the spectrometer in those places, I found two different wavelengths. My question is why does that happen?

Drakkith said:
Why do you have different colors if you're using a laser?
Well... I don't really know. With the spectrometer we tracked a blue light, if that helps. It was all very "robotic", which is why I'm having a bit of trouble explaining things here. We weren't given any insight on how exactly things worked.
 
  • Like
Likes mdbablu
  • #10
Techno_Knight said:
Well... I don't really know. With the spectrometer we tracked a blue light, if that helps. It was all very "robotic", which is why I'm having a bit of trouble explaining things here. We weren't given any insight on how exactly things worked.

Was the light from the laser blue? Was the blue light on the display of the spectrometer? Or something else?
 
  • #11
Drakkith said:
Was the light from the laser blue? Was the blue light on the display of the spectrometer? Or something else?
I assume the the source was a white light, but I'm not sure. I'm basing my assumption on the Professoer saying to "pick a color and track it". The Blue Light was the one we tracked with the Spectrometer. So it's not a case of putting a "blue tinted glass" or something in front of the Spectrometer. I wish I was ofmore help, but details about the experiment were scarce.
 
  • #12
Techno_Knight said:
I assume the the source was a white light, but I'm not sure. I'm basing my assumption on the Professoer saying to "pick a color and track it". The Blue Light was the one we tracked with the Spectrometer. So it's not a case of putting a "blue tinted glass" or something in front of the Spectrometer. I wish I was ofmore help, but details about the experiment were scarce.

By "steam of sodium" you don't mean a sodium-vapor lamp do you?
 
  • #13
Drakkith said:
By "steam of sodium" you don't mean a sodium-vapor lamp do you?
I... don't think so? To be completely honest, I don't know if it's Sodium Vapour or Mercury Vapour. The "Exercise Prepping" says it's Sodium, but the "Exercise Tasks" says it's Mercury. The Professor told us not to care about it, and the light source was inside a black box, so I don't know what it looked like.
 
  • #14
Techno_Knight said:
I... don't think so? To be completely honest, I don't know if it's Sodium Vapour or Mercury Vapour. The "Exercise Prepping" says it's Sodium, but the "Exercise Tasks" says it's Mercury. The Professor told us not to care about it, and the light source was inside a black box, so I don't know what it looked like.

Either way, those lamps are not lasers and do not have a single color, which makes sense now why you had to track the colors. After re-reading your posts, I must have thought the picture above with the laser source was your experimental setup and that you had used a laser. My apologies.
 
  • #15
Drakkith said:
Either way, those lamps are not lasers and do not have a single color, which makes sense now why you had to track the colors. After re-reading your posts, I must have thought the picture above with the laser source was your experimental setup and that you had used a laser. My apologies.
Oh don't worry about it. I'm coming off as a bit aloof, but the truth is I can't make much sense of this, as the leaflets were written around the mid 1900s, based on a book from the early 1900s. So I'm at a weird place where I don't understand everything from the experiment as it wasn't explained, the leaflet is incomplete, my book doesn't mention anything, I can't check the numbers because my Lab Partner has been AWOL for 2 days (technically I can, but I want someone else to run them again), and then there's the translation thing. Sorry for the whole mess is what I'm saying.
 
  • #16
I run the numbers again and I keep getting the same results. I checked the angles (θ & ψ) with the Professor while doing the Experiment, so these are right. I'm at a loss here. The wavelengths are bellow the optical spectrum, and I don't know why I get two wavelengths. The exercise says that we will get more than one, but I don't get the why.
 
  • #17
Techno_Knight said:
I... don't think so? To be completely honest, I don't know if it's Sodium Vapour or Mercury Vapour.
You can tell the two apart. Both are used in street lighting. Sodium is mostly yellowish and consists basically of one wavelength. Mercury is bluish-white and contains several different wavelengths (see below, sodium left, mercury right).
Lamps.png


Can you tell us if you looked closely at or were told what scattered the light? Was it a square grid as your picture implies or was it a linear array of many slits or something else? From your description,
Techno_Knight said:
... a small piece of glass with 570 grooves per mm and a... box, I guess?
it looks like you used a diffraction grating.

If you used a sodium lamp, then probably your task is to resolve the so called sodium D lines
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html

If you used a mercury lamp, then probably your task is to find the several wavelengths making up its spectrum
https://en.wikipedia.org/wiki/Mercury-vapor_lamp
 
Last edited:
  • #18
Upon doing some reading on the book, I happened upon this:

"If the light contains many wavelengths, then the maximum m-class for each wavelength corresponds to a certain angle"

Where m-class maximum refers to the formula dsinθlight = mλ m = 0,+-1, +-2 when talking about a Barrier of Diffraction (or like you called it Array of Diffraction Maxima).

So I think I found the answer to my question. What I called "fringes of confluence" are the m from above, and the book says that each one of them has its own wavelength.

As for the lamp, I did somedigging myself, and since I get wavelengths bellow 4k Angstrom, then it's probably a Mercury Vaporone, for according to the internet, it emits Ultraviolet light.
 
  • #19
Did you detect the fringes by looking at them or did you use a sensor of some kind? How did you measure the angles and how did you decide to label them ψ or φ?
You can find the Mercury spectrum here. Concentrate your attention on the "strong" lines, intensity of 1000 units because you are more likely to detect these before the others.
https://physics.nist.gov/PhysRefData/Handbook/Tables/mercurytable2_a.htm

By the way, the terminology for "m-class maximum" is "mth order maximum".
 
  • #20
We looked at them with the Spectrometer. As for the labeling that's... a bit complicated. Fringe 0 (m = 0) is φ, which we found as 50,5 deegres. Then we measured the next two fringes (+-1, +-2) and called them θ1/θ1' & θ2/θ2' (' goes to +, nothing goes to -). Then, from the θ angles we subtracted 50,5 and found ψ1/ψ1' & ψ2/ψ2' (now ' goes to -, and nothing goes to +). Then, using the formula tanφ = (sinψ - sinψ')/(2 - cosψ - cosψ') we found the "new"/real φ. It's a bit complicated, but those are the leaflet's instructions.

As for the Mercury Spectrum, I got λ = 1150 Angstrom for fringe 1, and λ = 3490 Angstrom for Fringe 2. I realize the first one makes no sense, but I run the numbers again, and that's what I get.
 
  • #21
I think I begin to understand what's going on here. Feel free to correct any misunderstanding I may have.
You used a diffraction grating to measure the spectrum of Mercury. Now the standard grating equation relating the wavelength to the angle at which the ##m##th order maximum is observed is ##a~\sin \psi=m~\lambda##. Here ##a## is the separation between grating lines. This equation predicts that the central, zeroth order maximum (##m=0##) occurs at zero angle. It also predicts that maxima of higher order (##m=1,2,3~...##) occur at equal angular displacements on either side of the central maximum, i.e. ψ = ψ'. You are not using this equation I think because you need to account for the possibility of a systematic error such that the central maximum is not exactly on the origin with respect to which you measure your angles. There is an offset φ corresponding to the angular displacement between the central maximum ##m=0## and your origin for measuring angles. Hence the equations ψ' = θ + φ & ψ = θ' - φ. In these expressions it looks like ψ' and ψ are the measured angles on either side of the central maximum, φ is the offset angle and θ is the true angle of deflection. For some reason that I don't understand, the equations allow for the possibility that θ ≠ θ'. Of course, φ must be a small angle. Your values of 69.6o and 32.9o are way too high.

Keeping all this in mind, try to find the offset φ. It's a number that when added to your measured ψ' and subtracted from the measured ψ gives the same result. What you get is small and consistent for the two measurements that you provided.
 
  • #22
That's pretty much it,yeah! Things that I should add:

The angles measured at the Lab are:

φ = 50,5 (m = 0)

θ1 = 70,9 (m = -1)
θ1' = 41,9 (m = +1)

θ2 = 85,2 (m = -2)
θ2' = 26,9 (m = +2)

I know that the results are pretty weird, but I did what the Professor said to do:

-Subtract φ from all the angles and get:

ψ1 = 20,4 (m = +1)
ψ1' = -8,6 (m = -1)

ψ2 = 34,7 (m = +2)
ψ2' = -23,6 (m = -2)

Then he told us to use tanφ = (sinψ - sinψ')/(2 - cosψ - cosψ') and find φ.

I know that the results are weird, but I don't know what else to do. That's what he told us. Do you have any ideas?
 
  • #23
An offset of 50.5o sounds too big. I can't imagine that your spectrometer was so grossly misaligned. Let me see what I can do with these numbers.
 
  • #24
kuruman said:
An offset of 50.5o sounds too big. I can't imagine that your spectrometer was so grossly misaligned. Let me see what I can do with these numbers.
Hate to say it but... yeah. There's another part of the exercise that we did there at the Lab (similar thing, but only for m = +-1, and with λ known, so we had to find a), and the offset 55,4! And that was overseen by the Professor. He checked all the values, angles, everything. I think it was supposed to be missaligned.

If you can find some more reasonable values I'd appreciate it! I tried all the formulas, but I still get the same results. I checked around and Mercury Vapor Lamps emmit UV light starting from around 180 nm. So I guess the 349 nm is valid enough, but it's the 115 nm that troubles me.
 
  • #25
There is something in all this that I do not understand, probably because I don't know the geometry of the experiment. If φ were a simple offset, then one should get equations 5 & 6 of the form sin(ψ±φ) = mλ/a. I don't see where the ±sinφ on the right hand side of the equations comes in. In any case, I got around it by adding equations 5 & 6 so that sin(ψ' - φ) + sin(ψ + φ) = 2nλ/a, which can be solved for the wavelength.

Again, not knowing how equations 5 & 6 were derived, I had to guess how to substitute the numbers. I used ψ'1 = 70.9o, ψ'2 = 85.2o, ψ1 = -41.9o and ψ2 = -26.9o in the above equation with two values for φ, one that you measured and one obtained using the formula provided by your professor, which is another thing I don't understand because I cannot derive it. I got 4 wavelengths, 2 first order and 2 second order that were close to each other and pretty close to one of the strong lines of mercury. I will not tell you which one - you have to discover it yourself.

Disclaimer
: you should understand that all this could be a whole bunch of numerology because I have not been able to ascertain the validity of the interpretation in terms of the geometry of the experiment.
 
  • #26
Hmm, okay, so I tried this:

I took this formula: λ = ([(sinψ + φ) - sinφ]/n)*a
a = 18300 Angstrom
φ = 50,5

For 1st Order: ψ = -41.9 & n = 1
So, with the angles being in their absolute values, we get λ = 4160 Angstrom

For 2nd Order: ψ = -26.9 & n = 2
So, λ = 1870 Angstrom

If I'm to use your formula, should I plug in the angles in their absolute values, or not?Apart from that, are these what you were talking about? To be honest, I don't know what's "correct" because like I said, he told us to use the formulas in the previous page in the manner I wrote down before. I looked in the internet about Mercury Vapor lamps and found out that they emit a UV light, so I guess bellow 4k Angstrom is an acceptable wavelength, right?
 
  • #27
Techno_Knight said:
I took this formula: λ = ([(sinψ + φ) - sinφ]/n)*a
I didn't have much luck with this formula probably because I don't understand how to use it. If I could derive it, I would be able to identify which measured variable is what in terms of the geometry. For that reason, I decided to add equations 5 & 6 hoping that whatever misconceptions I had will cancel out. The consistency of the results indicates that it probably did. However I still don't understand equations 5 & 6, hence my disclaimer.

Regardless of what wavelengths Mercury lamps emit, did you or did you not see with your own eyes the m = 1 and m = 2 maxima? If you did, what color were they? If you didn't, what did you use to detect the maxima in the UV range? Your eyes are insensitive to UV or IR radiation so, if you saw with your own eyes the maxima, wavelengths outside the visible range are out of the question.
 
  • #28
When a diffraction grating is tilted at an angle of ## \theta_i ##, the equation for constructive interference is ## m \lambda=d (sin(\theta_i)+sin(\theta_r) ) ## with ## m= ## integer, and where the angles are both measured relative to the normal to the grating. Laboratory experiments with diffraction gratings can be quite interesting, but for this particular experiment, the available information is too limited to be able to make heads or tails of it, and they have you using too many formulas without any explanation of where they came from. ## \\ ## One additional note: If the source was non-monochromatic=e.g. somewhat white, you can always tell the m=0 location, because that location will be the same color as the source=white. For a transmissive grating, its where the light goes straight through, and for a reflective grating, it's where angle of incidence equals angle of reflection. There is a sign convention in the above formula in what direction is positive for the angles, so that the m=0 condition has the two terms canceling each other on the right side. ## \\ ## Normally it helps to be able to physically measure any tilt angle on the diffraction grating, rather than trying to treat it as an unknown.
 
Last edited:
  • #29
kuruman said:
I didn't have much luck with this formula probably because I don't understand how to use it. If I could derive it, I would be able to identify which measured variable is what in terms of the geometry. For that reason, I decided to add equations 5 & 6 hoping that whatever misconceptions I had will cancel out. The consistency of the results indicates that it probably did. However I still don't understand equations 5 & 6, hence my disclaimer.

I searched around a bit more, and the leaflet says that they worked according to the Alonso-Finn model to come up with the two equations.

kuruman said:
Regardless of what wavelengths Mercury lamps emit, did you or did you not see with your own eyes the m = 1 and m = 2 maxima? If you did, what color were they? If you didn't, what did you use to detect the maxima in the UV range? Your eyes are insensitive to UV or IR radiation so, if you saw with your own eyes the maxima, wavelengths outside the visible range are out of the question.

I saw a blue light, at m = +-1 & m= +-2, so I guess the UV thing is out.

From your equation:

1st Order | φ = 50,5 | ψ & ψ' in their real values | a = 1800 Angstrom | m = 1
λ = 0,4558 μm = 4558 Angstrom

2nd Order | φ = 50,5 | ψ & ψ' in their real values | a = 1800 Angstrom | m = 2
λ = 0,4436 μm = 4436 Angstrom

Now, if I transform φ due to the Professor's equation, I get:

For ψ1 & ψ1': φ = -60,1
For ψ2 & ψ2': φ = -54,7

And again:

1st Order | φ = -60,1
λ = 1,59 μm = 15900 Angstrom

2nd Order | φ = -54,7
λ = 0,747 μm = 7470 Angstrom

Are any of these right?I have to be honest, I'm pretty lost here. I get what you're saying, more or less, but I'm really not sure since I didn't really "get" the exercise. I think I'll just turn in the exercise with the original numbers, and ask him later where the problems lie.

There was another part of the exercise, which was the same thing, but with Sodum Vapor, and instead of wanting λ, we were looking for a (the one we're using now). We solved it the same way I found my (illogical) wavelengths (the ones which were <4k), so I guess I'll leave the results as is and ask later. I don'tcare about the grades anyway, someone like me is never going to get a 9/10 or graduate with a 4.0 GPA.

Charles Link said:
When a diffraction grating is tilted at an angle of ## \theta_i ##, the equation for constructive interference is ## m \lambda=d (sin(\theta_i)+sin(\theta_r) ) ## with ## m= ## integer, and where the angles are both measured relative to the normal to the grating.

Hmm, I've never seen this one, but I get the meaning behind it. It's the usual mλ = dsinθbright where m = 0, +-1, +-2, ... formula, but with the tilted angle taken into account.

Charles Link said:
Laboratory experiments with diffraction gratings can be quite interesting, but for this particular experiment, the available information is too limited to be able to make heads or tails of it, and they have you using too many formulas without any explanation of where they came from. ## \\ ## One additional note: If the source was non-monochromatic=e.g. somewhat white, you can always tell the m=0 location, because that location will be the same color as the source=white. For a transmissive grating, its where the light goes straight through, and for a reflective grating, it's where angle of incidence equals angle of reflection. There is a sign convention in the above formula in what direction is positive for the angles, so that the m=0 condition has the two terms canceling each other on the right side. ## \\ ## Normally it helps to be able to physically measure any tilt angle on the diffraction grating, rather than trying to treat it as an unknown.

Yeah, I found that bit about m = 0 being white, due to all the different colors of the source's non-monochromatic light meeting, over at my book.
 
  • #30
Techno_Knight said:
... 570 grooves per mm ...
I believe your value for a is incorrect.
570 grooves/mm = 570 grooves/mm × 1000 mm/m = 5.7×105 grooves/m. Therefore a = 1/(5.7×105) m/groove = 1.75×10-6 m/groove.
Multiply by 1010 Angstrom/m to get a ≈ 18000 Angstroms/groove. This will increase your wavelengths by a factor of 10.

Also, can you show a sample calculation of how you got the values that you got? They disagree with mine.

On edit: I agree with the first two calculations even though it appears you have the wrong value for a. I disagree with the other two after "And again". You might wish to consider doing these on a spreadsheet with the same formula for all. Then either all of them will be correct or all wrong but easy to trouble shoot.
 
Last edited:
  • #31
kuruman said:
I believe your value for a is incorrect.
570 grooves/mm = 570 grooves/mm × 1000 mm/m = 5.7×105 grooves/m. Therefore a = 1/(5.7×105) m/groove = 1.75×10-6 m/groove.
Multiply by 1010 Angstrom/m to get a ≈ 18000 Angstroms/groove. This will increase your wavelengths by a factor of 10.

a is from the previous exercise I mentioned, which we did at the Lab, and checked it. Basically we did the same thing as here, but with Sodium Vapor, known wavelength (λ = 5893 Angstrom), for just one fringe (m=+-1) and using the formula for φ (tanφ = ...) and either (5) or (6) found a. The Professor checked it and said it was correct, so I have to use it here.

I don't think it's out of the question that all these values are incorrect, but since he told us to treat the exercise a certain way, I'll leave the results as is and ask him later. Just an FYI, he's a Post-Grad/Junior Professor so he might be glossing over some stuff to get it done quickly.
kuruman said:
Also, can you show a sample calculation of how you got the values that you got? They disagree with mine.

On edit:
I agree with the first two calculations even though it appears you have the wrong value for a. I disagree with the other two after "And again". You might wish to consider doing these on a spreadsheet with the same formula for all. Then either all of them will be correct or all wrong but easy to trouble shoot.

I just took the equation you came up with, and plugged in the angles, in the real values, not the absolute ones. For the 1st Order I put m = 1, and for the 2nd m = 2.

1st Order:
sin(70.9 - (-60,1)) + sin(-41,9 + (-60,1)) = 2*1*λ/(18300*10-10 Angstrom) <=> λ = -2044 Angstrom

2nd Order:
sin(85,2 - (-54,7)) + sin(-26,9 + (-54,7)) = 2*2*λ/(18300*10-10 Angstrom) <=> λ = -1579 Angstrom

Huh. Those are obviously different from before. I honestly don't know why. Maybe I added them in their absolute values before? Are these anywhere close to reality?
 
  • #32
Why are you using negative values for φ? I didn't. Like I wrote earlier, having the picture, on which the derivation of equations 5 & 6 was based, would help tremendously with the correct substitution of variables.
 

Related to Barrier of Diffraction: Different Wavelengths, Same Color?

1. What is the barrier of diffraction?

The barrier of diffraction refers to the phenomenon where light waves of different wavelengths, but the same color, are diffracted differently when passing through a narrow opening or around an obstacle. This results in the separation of the wavelengths and the formation of a diffraction pattern.

2. How does the barrier of diffraction affect the perception of color?

The barrier of diffraction can affect the perception of color by causing a separation of the wavelengths, leading to the perception of different colors. This is particularly noticeable when white light is diffracted, resulting in the appearance of a rainbow of colors.

3. What is the role of wavelength in the barrier of diffraction?

Wavelength plays a crucial role in the barrier of diffraction as it determines how much a light wave will bend when passing through a narrow opening or around an obstacle. Longer wavelengths, such as red, are diffracted less than shorter wavelengths, such as blue, resulting in the separation of colors.

4. Can the barrier of diffraction be observed in everyday life?

Yes, the barrier of diffraction can be observed in everyday life. For example, when light passes through a small opening in a window blind, it can create a diffraction pattern on the wall. Diffraction can also be seen when light passes through a CD or DVD, resulting in the separation of colors.

5. How is the barrier of diffraction used in scientific research?

The barrier of diffraction is used in scientific research to study the properties of light and to analyze the structure of materials. It is also used in various optical instruments, such as spectrometers, to separate different wavelengths of light and measure their intensity, which can provide valuable information about the composition of a substance.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
Replies
1
Views
2K
Replies
5
Views
2K
Replies
2
Views
968
Replies
4
Views
434
Replies
3
Views
940
Replies
66
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
618
Replies
21
Views
4K
Back
Top