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

[MAULIN]

is it necessary to get 4 diffracted beams (excluding incident beam) as shown in the picture on this site???

[deleted-71712]

Hi Maulin,

Welcome to the forum!

Calculating the spacing from multiple diffracted beams is like measuring anything multiple times -- it reduces the error in your final result. This error could be something like plus or minus a degree from reading the protractor, or something more seriously like thinking you were seeing m = -1, 0, and +1 when you were seeing m = 0, +1, and +2. So yes, you can calculate the spacing with fewer diffracted beams, but it might be a somewhat less accurate. One thing you could do to compensate is change theta_i, the angle of the incident beam -- the result for d should be the same with the set of diffracted beams visible at a different angle.

Hope that helps, and let us know if you have more questions,
Amanda

(for reference to others reading the thread, the experiment is here: https://www.sciencebuddies.org/science- ... p011.shtml)

[MAULIN]

thank u............amanda!!!!

[MAULIN]

i m doing project on LASERS.........
can any one help me in getting a good 'conclusion on lasers' for my prjct????

[deleted-71712]

Hi Maulin,

You can find general info about writing a conclusion here: https://www.sciencebuddies.org/science- ... ions.shtml

and a guide to the whole science project process here: https://www.sciencebuddies.org/science- ... ndex.shtml.

A conclusion is something that you write after you've finished doing your experiment; if your "just starting" status is not outdated, it's probably a bit soon to worry about this part. If your teacher is asking for a draft of your paper, perhaps you could outline the kind of information you plan to put in the conclusion and give an example like, "If [example result] is observed, then I could conclude that [example conclusion]."

Good luck,
Amanda

[MAULIN]

actually i wntd a readymade 'conclusion on lasers' if possible..........i hv to submit my prj. in few days!!!

[MAULIN]

i m nt able to get clear diffracted rays as shown in the diag.(in this site)......
i m using a red laser pointer with wavelength of 630-680 nm........is this nt enough???
any other factors which i need to keep in mind?????

[deleted-71447]

Hi Maulin,
I combined your 3 threads into one. Please keep future comments about this experiment in this same thread so that we can figure out what you are doing and provide the best help possible.

Regarding providing a ready-made conclusion, sorry, but we can not do that. As Amanda mentioned, the conclusions are written based on your hypothesis and the results of your experiment. If/when you have specific questions about how to synthesize your hypothesis and experimental results into a conclusion, then please provide some details about your hypothesis and results, and then we can probably help.

Regarding the problems you are having with the experiment, do you see any diffracted beams at all? I don't see any problem with the laser you are using. Are the batteries low? The photo on the experiment page has been altered to make the diffracted beams more distinct and obvious, so what you see will not be as clear as what is shown in the photo. Does it help if you move to a darker room where you can see the rays more clearly? If it still seems not to work, you can try to make small adjustments to the experimental setup to see if anything helps to produce clear refracted rays. If you're still stuck, let us know!

Good luck,
chris

[MAULIN]

thnx for the help!!

[deleted-71447]

You're welcome. Good luck with the rest of the experiment.

[MAULIN]

i conducted the experimnt on a std. CD........
i got the value 1883.515333 nm
is it acceptable??

[deleted-71712]

Hi Maulin,

To start, I will tell you that your result is definitely in the ballpark -- well done! Your procedure and calculations are very likely done correctly, although it's possible that you could change some things to make your measurements a bit more accurate. One thing you should be careful of when reporting results is significant figures, a concept that's related to error. For example, if you reported your result as 1883.5 nm (with no +/-), you would be implying that you're pretty sure the value you measured is closer to 1883.5 than 1883.4 or 1883.6 nm. On the other hand, you might decide to report it as 1900 nm, meaning that it's closer to 1900 than 1800 or 2000 nm.

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

It's even better to explicitly state the error, so you might end up with something like 1883.5 nm +/- 250 nm -- in which case you don't lose much by rounding to 1880 nm +/- 250 nm, because the last few sig figs don't mean much with that much uncertainly. 250 nm is just an example -- you need to estimate the sources of error in your experiment, such as from reading the angle on the protractor, and then propagate them to your final result.

http://en.wikipedia.org/wiki/Measurement_Uncertainty
http://en.wikipedia.org/wiki/Error_propagation

It's also worth googling both of those terms and checking the indices of your science books for different explanations. If the error-propagation math is too heavy, you can also do this: Say you measure the angle as 60 degrees +/- 2 degrees. Do the calculations using 58, 60, and 62 degrees. Your result is the one you got using 60 degrees, but you want to choose an error that sets the upper and lower bounds near the results for 58 and 62 degrees, respectively. It's a little more complicated if your result is the average of calculations from several different reflected beams, but you can at least do this for each calculation individually. If the errors are the same for each beam, the general rule of thumb is that the error in the average will be the error in one measurement divided by the square root of the number of measurements that went into the average -- but you should look this up, try to follow the reasoning behind it, and cite a good source (e.g. a statistical handbook) when you turn in your project.

At this point, you should have a decent idea of how much inaccuracy there is in your result due to error. Don't be discouraged by this -- there is error in every single scientific experiment, and it's impressive that you can measure such a tiny length scale with things around the house! Now, you want to look up the actual value. If it falls within your error bounds, then it means that your result is probably as good as it could be within the limitations of your measurement tools. If it's outside of your error bounds but still pretty close -- i.e. the same order of magnitude -- then it's likely that there's a source of error that you've neglected. If it's wildly different, then you should first check your calculations and then carefully repeat the experiment if necessary.

It's a good idea to estimate the error first, because then you won't be tempted to choose larger than reasonable error in your degree measurement, etc, just to make it work out. And if your 'blind' measurement of the track spacing is accurate within your error bounds, you really have something to be proud of. When you're ready, the track spacing can be found in this article:

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

(As a side note, the Greek letter mu, like a u with a funny tail going down in front, is used as a metric prefix meaning micro-, since m and M are taken for milli- and mega-. Micrometers are also called microns, perhaps because there are instruments called micrometers, though the stress is on the second syllable for the instrument when spoken aloud. Just a heads-up since it took me years to get this straight.)

Best wishes, and let us know how it turns out!

Amanda

[hook93]

hello everyone, i am doing this project for my science fair. I finished the research phases, but now i have the experiment to do which is due jan 18. I have some questions about the project that i didnt understand. First thing is that how do i know which defracted angles are m=1, m=-1 etc.....,,, second thing, i dont get how to replace in the formula the stuff that i get, they say its like this
d = m × λ / (sin θm − sin θi ) but what should i put in sin theta-m,, i only put the angle of the defracted beam such as the angle of m=-1?? or should i also multiply the angle by -1 i know at the m part at the beginning of the formula i should put for example m=-1... and last thing is that when i get the measurements for the angles, i have to do the formula for each measurement and then find the average of all of the d's found??

[kgudger]

Hello and welcome to the forum!

If you look at the experiment figure under "making measurements", you will see the various angles labeled. Please note that the m=0 reflection is the one where the angle of reflection = the angle of incidence. Angles on the same side as the reflected beam are positive m, those on the incident side are negative m.

With regards to the formula, the example states:

For example, for m = −1, and a laser pointer with a wavelength of 655 nm, the formula would be: d = (−1) × 655 / (sin θ−1 − sin θi )

. If you set up your calculator to use degrees for trigonometric functions, then θi is the incident angle in degrees, and θ−1 would be the m=-1 angle in degrees.

With regards to your final question, Yes, you should do the formula for each measurement and

Calculate the average value for each d column, and, separately, for all of the values of d.

.

I hope this helps!
Keith

[hook93]

thank you for your help, but i have another question. Concerning the values that i replace to then find the average of the D's found, i shouldnt replace the m=0 value right,?? since the result would be 0 anyways?? or should i?

[kgudger]

Hi:

For each different CD, and each trial, you should measure the incident angle, and use that value for the calculations across the data row in the table. You don't calculate 'D' for the incident angle, as the sin θi - sin θi will always be zero. In your actual experiment, you probably do not change the placement of the laser, so your incident angle will probably always be the same. However, a good scientist will make sure and measure it each time anyway! Good luck.

Keith

[hook93]

thank you for your help, if i need anything ill post again...

[hook93]

ok, i did the experiment and it worked fine. But the results that i got are a bit unbelievable. i did 3 different trials. One trial on a dvd that had information, another one was on an empty dvd, and the other was on a CD that had information. I got 1560nm for the empty dvd and 1703nm for the CD with information, and i got 1434nm for the dvd that had information. Are these results ok???

[hook93]

and by the way, can you have more than 5 reflected angles.??

[elykl33t]

Hi

I'm doing my science project on this topic. When I use a CD, I get 4 clearly defined refraction beams, as shown on this site. However, when i use a DVD I can only get two refraction beams no matter what angle I put the incident beam at. Is this normal? How do you know which two beams they are? (m=1,m=0,m=-1,etc.)

Thanks!!!!

[kgudger]

Hello and welcome to the forum.

I haven't done this experiment myself, so I don't know if DVDs normally give you only 2 refracted beams. You say you've tried several different incident angles, which would be the most logical thing to try. How dark is the room you're working in? For the DVD, with closer together tracks, you might need to have the room darker to see the beams. You also might try changing the distance from the DVD to the laser.

In terms of which 'm' the beam is, in the experiment it shows that beams on the other side of the perpendicular from the incident beam are positive, with the beam with the same angle as the incident beam = m=0. As you move away from the perpendicular on the same side as the incident beam, the first beam is m=-1 and so on. HTH.

Keith

[kgudger]

WIth regards to this question:

I got 1560nm for the empty dvd and 1703nm for the CD with information, and i got 1434nm for the dvd that had information.

I searched the web for an answer, and it looks like the CD number is reasonable. However, DVD tracks are less than half the distance as CDs apart.

With regards to this question:

and by the way, can you have more than 5 reflected angles.??

It would seem that you've done a great job getting the experiment to work! Congratulations!

Keith

[elykl33t]

I have made sure the room i'm working in is dark, however i can still get only 3 refraction beams. My problem is that while i know one of them is m=0, i can;t be sure whether the one on the left is m=1 and the one on the right is m=-1. What do you think?

Also, a quick question regarding the formula to calculate this. Say you get a formula like 2X650\(75-5)....you get about 18.57, not 1857 which is an "in the ballpark" answer. Are you supposed to move the decimal like that? why?

[deleted-71487]

Say you get a formula like 2X650\(75-5)....you get about 18.57, not 1857 which is an "in the ballpark" answer. Are you supposed to move the decimal like that? why?

Values of the sin function are always in the range 0-1, so I'm not sure where you got (75-5) from. Did you forget to take the sin of the angles?

[deleted-71588]

You first need to identify the m=0 beam. It should be on the other side of vertical with the same deviation. In the project photo, the incident angle is 70 degrees and the m=0 is at 110 degrees (70 + 110 = 180). The m=+1 will be the beam with the next higher angle, example shows 138 degrees. m=-1 will be at the next lower angle, example shows 84 degrees. m=-2 will be at the next lower angle, example shows 56 degrees.

As far as the formula, you are doing something wrong in the denominator. The range of the Sin function is -1 <= (Sin x) <= +1 for all angles x so the domain of the denominator is -2 <= denominator <= 2. Did you forget to use the Sin function and simply try to use the angles directly?

If you aren't getting m=+1, m=0, m=-1, and m=-2 beams, have you tried using different incident angles? Is your incident beam on a "radial" (co-planar with a radius line of the CD/DVD)? You should be able to get these to show up.

[elykl33t]

Oh thanks guys! sorry that was just stupid of me. I forgot how sin and all that works, and i just assumed "sin1" or w/e meant the angle of the beam. Yah i get it now.

And so basically, since i see one beam left of m=0 and one right of it, those are m=1 and m=-1 respectively?

[deleted-71588]

since i see one beam left of m=0 and one right of it, those are m=1 and m=-1 respectively?

That would be my expectation; however, make sure you aren't being confused by the incident beam. There is a possibility that you chose an incident angle so the incident beam and the m=-1 or m=-2 are at the same angle which could cause considerable confusion.

[elykl33t]

PLEASE HELP!!

the formula on the site, d = m × λ / (sin θm − sin θi ), does not make sense to me!!!!

So, the "m" is the value of m (1, -1, -2, etc.) i understand that

the weird A shapedish thing is wavelength of the laser pointer i understand that

now, "sin θm" refers to the sin of the angle of your given beam correct? so if you did m=1, and its angle was +55 degrees, itd be sin(55) right?

and then sin θi refers to the sin of your incident beam, correct? so if the incident beam is 20 degrees, its the sin(20)?

my problem is that, as an example problem, if you do 1 X 650/(sin(50)-sin(20)), its comes out negative!!!!!!!!! i've done this myself multiple times, and it wont work! also, this is nearly identical to what it would be using the picture on this site correct????? PLEASE HELP!!!!!!!

[deleted-71588]

I don't understand how you get a negative. Sin 0 = 0, Sin 90 = 1, between 0 and 90 degrees, Sin is a monotonic increasing function. So Sin 55 > Sin 20 so the denominator is positive. Could you be using a calculator set up for calculating in Radians instead of Degrees?

Caution: The angles in the equation are supposed to be measured from "normal" or perpendicular to the CD/DVD surface.

Note: For the case of m=-1, the denominator is expected to be negative to cancel out the negative in the numerator.

[elykl33t]

Yes! that was the problem, thank you. I didn't even think of that.