Using a Laser Pointer to Measure the Data Track Spacing on C

[fsbs7983]

Hi guys:
So I decided to do this project for my science project. Hooray. After conducting one of my tests the other night and measuring the angles, I became very confused. How can I tell which beam is the incident beam, and which is the reflected beam? Is the incident beam the closest one to the original laser beam, or what? I understand that m=+1 and m=-2 are different orders, but how can you label them correctly if you dont know where the incident beam is?
And by the way, WHAT IS θm? The project just says that θm is the angle of the Mth diffracted ray...............WHAT DOES THAT MEAN??????

HELP PLEASE!

Please respond ASAP so I can finish my project on time. Thanks
-fsbs7983

would the incident beam be the longest beam? and i noticed that with my disc, there are 2 beams right of the 90 degrees and 3 to the left, but on the project in the pic, there are 3 on the right and 2 on the left. IS THIS A PROBLEM?

[deleted-71712]

Hi fsbs7983,

Welcome to the forum. I've moved your post into a new topic to make sure any experts who respond will realize that you're a new poster, but here are links to the project description and the previous discussion that you found:

https://www.sciencebuddies.org/science- ... p011.shtml
http://sciencebuddies.com/science-fair- ... =29&t=3955

To answer your first question, look at the photo under step 2a. The angle of incidence is always equal to the angle of (zeroth-order) reflection (http://en.wikipedia.org/wiki/Reflection_%28physics%29), so theta_0 = theta_i. The incident beam is the one coming from your laser pointer.

About angles... have you studied trigonometry at all -- sines, cosines, tangents? If not, this is a good place for some basic information, and you could also check your math textbook:

http://en.wikipedia.org/wiki/Angle

Amanda

[fsbs7983]

Thanks Amanda for your quick reply.

I apologize, I should have done more research on angles beforehand so I could understand different terms and laws about light and diffraction. I understand the incident angle a little better now. Anyways, when we are shining the laser beam down the face of the card, are we shining it straight down the card, aligned with 90 degrees? Or does the angle at which the beam is pointed not matter as long as the beam points at the center of the protractor? I ask this because in the picture (2a) the incident is way off from the 90 degrees while here I thought our incoming beam should be lined up with the 90 degrees. Does varying the angle of incidence affect the credibility of our results? I don't think it should (maybe the question was answered before), but I just want to make sure.

A problem I encountered, was that if we point the laser straight down 90 deg. on the protractor, theta_m is abnormally small. It is difficult to discern the placement of the reflected beam when doing this because it may be shining back at your hand or even on the laser pointer!
By the way, when measuring our angles, say theta_i is 26 deg. and another angle is 22 deg. Since one's hand shifts often when marking the beam placement on the card, and the angles may come out slightly under/overestimated, is it safe to assume that the 22 deg. angle is theta_m if each of the diffracted beams are spaced a fair distance away from each other?

And to answer your question, no, I haven't dealt with trig. yet. I understand how to calculate track spacing, I just don't get all the fancy science behind sin, cos, and tan. I didn't think I needed a full background of trig. to at least start this project.

Thanks again for your time
fsbs7983

[deleted-71588]

are we shining it straight down the card, aligned with 90 degrees? Or does the angle at which the beam is pointed not matter as long as the beam points at the center of the protractor?

Hey, this is science. You asked a couple of questions that you can probably quickly experiement with and find out your own answers. In order to accurately measure an angle with a protractor, the vertex of the angle must be alligned with the center of the protractor's arc. This is basic how do you measure an angle with a protractor. You could figure that out by measuring the inside angles of a triangle and adding them up and seeing how close you get to 180 degrees. If you offset the protractor the same direction at each vertex, does it add up to more or less than 180. Offset the protractor the opposite direction and again what do you get.

Try different incident angles. Find one which produces the most easily visible reflected angles, do the calculations. Choose another incident angle, do the calculations. Choose yet another and calculate. See how close your answers come out.

[fsbs7983]

Thanks Craig for your input.

I have a pretty good hunch I already know the answer to my question but I'll try your methods to be sure.

[fsbs7983]

Should the number of diffracted beams vary between different incident angles? When I shine down 80 deg.(counting from left to right) I see only 4...is it possible that a beam may be diffracted close to the cardboard spacer @ 160-180 deg (left to right)?

I ask this because at other angles I saw 5 beams... I understand you need only 3 beams to measure pitch, but I'm wondering... if you happen to see more than 3, you should calculate those to get the best results, correct?

~fsbs7983

[fsbs7983]

Ok, calculating results.

When using the equation, I encountered 2 different answers for the same numbers. I also got numbers "out of the ballpark"
Here's what I did:
1. -1x650/sin(-7⎯sin(21= 5075.151352
2. -1x650/sin(-7)⎯sin(21)=5333.222513
?????????????????????????????????????????????????????????????????????????????????????????
Why did adding an end parentheses change the answer? Confused please help. And why are my answers so off?

Heres my data so far: wavelength 650
Trial 1: CD theta_i= 10; theta_1=40; theta_2= n/a; theta_-1=-40; theta_-2= n/a
d, m=1 (nm) 1014.9; d, m= -1= 1007.6

Trial 2: CD theta_i= 13; theta_1=40; theta_2=80; theta_-1=-52
d, m=1 (nm)= 1016.0; d, m=2 (nm)= 1321.0; d,m=-1 = 822.3

Trial 3: Cd theta_i=21; theta_1=47; theta_2= n/a; theta_-1=-7; theta_-2= -34; theta_-3= -69
(the calculations are the ones at the top)

[fsbs7983]

Hello? Is anyone available? I need help as soon as possible please!

[deleted-71712]

In both of your calculator entries, you're missing a set of parentheses around the denominator of the fraction -- these are the only parentheses that appear in equation 2, step 3b, here:

https://www.sciencebuddies.org/science- ... p011.shtml

But don't get rid of the parentheses surrounding the arguments of the sine functions -- these aren't written in equation 2 but will be required for entry in virtually any calculator.

Compare your current answer: http://www.google.com/#sclient=psy&hl=e ... 20degrees)

To the answer you get when including parentheses around the denominator: http://www.google.com/#hl=en&sugexp=gsi ... +degrees))

The other thing to watch out for is whether your calculator is in radian or degree mode -- but yours seem to be in degree mode, which is what you want.

Amanda

[deleted-71588]

Should the number of diffracted beams vary between different incident angles?

Yes! This is something I expected you to discover yourself.

When I shine down 80 deg.(counting from left to right) I see only 4...is it possible that a beam may be diffracted close to the cardboard spacer @ 160-180 deg (left to right)?

Maybe, but doubtful. The wave equations for diffraction gradients is a set of four complex equations (as in equations involving complex numbers - real and imaginary parts) attributed to Maxwell and Schrodinger. These are well beyond what most BSEE even want to think about trying to solve after 30 hours of higher math (calculus, differential equations, and transforms).

Each of the visible rays is in fact a "real root" (polynmial factor) to a solution of these equations. Because the solutions can be both real and imaginary, the most likely explainations are that the ray that disappeared for your 80 degree incident angle was an imaginary solution or a conjugate pair of one of the other rays.

If you want to play with some numbers, try working the simplified equations backwards using the expected values you calculated and plugging in different ordinals than the ones you observed and find out what angle the ray for that ordinal would be expected to be located at.

[fsbs7983]

started fixing some of my calculations, but I still found a sketchy answer:

http://www.google.com/#hl=en&sugexp=ldy ... d84f286047

any suggestions?

another weird answer:
http://www.google.com/#hl=en&sugexp=ldy ... d84f286047

and yet another:
http://www.google.com/#hl=en&sugexp=ldy ... d84f286047

Should I just redo my trials with better angles? Please help!

fsbs7983

[deleted-71588]

The links (URL's) you posted appear to be some dynamic google search results and what I'm seeing is probably not what you were looking at when you posted them.

[fsbs7983]

Really? Because when I look at them they are fine.
well this is what i put in the google search bar: (in order of the links posted earlier)
another thing, I noticed that I only had this problem with m=-1.

650*-1/(sin(-52 degrees)-sin(13 degrees))
answer:
(650 * (-1)) / (sin(-(52 degrees)) - sin(13 degrees)) = 641.682633
-----------------------------------------------------------------------------------------------------------------

650*-1/(sin(-40 degrees)-sin(10 degrees))
answer:
(650 * (-1)) / (sin(-(40 degrees)) - sin(10 degrees)) = 796.143445
--------------------------------------------------------------------

650*-1/(sin(-38 degrees)-sin(10 degrees))
answer:
(650 * (-1)) / (sin(-(38 degrees)) - sin(10 degrees)) = 823.504435
------------------------------------------------------------------------------------------------

[deleted-71588]

You need to go back and look at the picture under Making Measurements part 2 in the write up https://www.sciencebuddies.org/science- ... p011.shtml again.

Your m=-1 assumptions yielded values that are too close to the wavelength of the laser. If the tracks were actually spaced that close, then a laser of that wavelength (or the ones used in the readers) couldn't reliably read the data. I built an Excel spread sheet and entered your data:
650 -1 -2 -3 -4
-52 13 641.683 1283.365 1925.048 2566.731
-40 10 796.143 1592.287 2388.430 3184.574
-38 10 823.504 1647.009 2470.513 3294.018
Even an m=-2 assumption would yield track spacing that is too closely spaced to read with lasers. If you confused which side of 90 the angle was on and these are really some positive m values:
600 1 2 3 4
52 13 1154.407 2308.814 3463.221 4617.628
40 10 1385.516 2771.031 4156.547 5542.062
38 10 1470.544 2941.088 4411.632 5882.176
an m+1 assumption yields too closely spaced tracks.

You need to measure ALL the reflection angles for each incident angle you try. Having ALL the values and the correct sign indicating which side of 90 degrees the angle was actually on should allow figuring out which m value corresponds to which reflection angle because the results for all reflection angles with the proper assumption of m angles should yield close to the same values.

[danielpower]

Thanks for your tip,it helped me a lot about the information of laser dazzler.