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CD Track spacing
Posted: Mon Feb 19, 2007 7:30 pm
by GRAHAMRICHARDS
Hello
My son, Grade 7 wanted to do this experiment for his science project. I thought I could do the math! But I am getting an answer 10x the 500 nm that we are looking for. I’ve reread the instructions and have checked the reflected angles with my son. Below is the web page and the formula. I hope someone can help.
https://www.sciencebuddies.org/mentorin ... ?from=Home
d = (−1) × 655 � (sin θ−1 − sin θi )
Posted: Mon Feb 19, 2007 7:44 pm
by deleted-71576
Could you please write down all the steps on the math you are doing.
We'll take a look at that first, to make sure it's not just a math error. If not, let's look more closely at what you did.
But first, let's make sure the math is correct.
Posted: Tue Feb 20, 2007 1:26 pm
by GRAHAMRICHARDS
here's the equation
d = (−1) × 655 � (sin θ−1 − sin θi )
d=-1X655/(sin 15- sin 20)
=7872.49 nm
thank you
Posted: Tue Feb 20, 2007 6:51 pm
by deleted-71576
Ok, one of the other experts here reviewed the data, and also agrees that the math seems fine. To summarize her thoughts:
There seem to be three possibilities:
1) The angle is totally wrong (mismeasured)
2) The diffraction order is misidentified
3) The laser wavelength is wrong (655 is the default in the instructions)
Can you give us the the entire data set for all diffraction orders to see if something jumps out at us there.
Meanwhile, check the laser wavelength and make sure that you have identified the correct diffraction order.
You should look over this post to see if a light bulb goes off:
https://www.sciencebuddies.org/mentorin ... php?t=2247
I had originally thought that the formula might require radians instead of degrees. I'm still mulling over that one.
CD Track spacing
Posted: Tue Feb 20, 2007 7:16 pm
by GRAHAMRICHARDS
Again I'm looking for an anwer of 500 nm
thanks
calculate laser pointer wavelength as follows
630+680 = 655 nanometres
2
I checked the angles with my son again and still getting answer of 2-4 Um.
Here's the angles. Note didn't use m=0.
Incident angle 20 (70 on protractor)
+1 30 (60 on protractor)
+2 50 (40 " ")
-1 15 (75 " ")
-2 0 (90 " "}
Incident angle 30 (60 on protractor)
+1 57 (43 on protractor)
+2 75 (15 " ")
-1 30 (60 " ")
-2 10 (80 " "}
Incident angle 40 (50 on protractor)
-1 30 (60 " ")
-2 3 (87 " "}
Posted: Tue Feb 20, 2007 7:17 pm
by Louise
zzzzdoc wrote:Ok, one of the other experts here reviewed the data, and also agrees that the math seems fine. To summarize her thoughts:
There seem to be three possibilities:
1) The angle is totally wrong (mismeasured)
2) The diffraction order is misidentified
3) The laser wavelength is wrong (655 is the default in the instructions)
Can you give us the the entire data set for all diffraction orders to see if something jumps out at us there.
Meanwhile, check the laser wavelength and make sure that you have identified the correct diffraction order.
You should look over this post to see if a light bulb goes off:
https://www.sciencebuddies.org/mentorin ... php?t=2247
I had originally thought that the formula might require radians instead of degrees. I'm still mulling over that one.
It should not matter if you are in radians or degrees as long as your calculator is set properly. That is: sin (90) in degrees will be the same result as sin (pi/2) in radians.
Louise
Posted: Tue Feb 20, 2007 7:31 pm
by deleted-71576
I agree. I was wondering (out loud) what the result would be if the calculator was set improperly.
Posted: Tue Feb 20, 2007 7:43 pm
by Louise
zzzzdoc wrote:I agree. I was wondering (out loud) what the result would be if the calculator was set improperly.
Sorry, I misunderstood what you were saying...
The "answer" should in fact be 700 nm or something like that for the track spacing. Cannot find the link now. It is a bit larger than the 500 nm quoted by Mr. Richards, but obviously not a factor of 10.
Louise
Re: CD Track spacing
Posted: Tue Feb 20, 2007 7:45 pm
by Louise
GRAHAMRICHARDS wrote:Again I'm looking for an anwer of 500 nm
thanks
calculate laser pointer wavelength as follows
630+680 = 655 nanometres
2
Why are you averaging these two numbers? This isn't the source of the error, but the laser has one fixed wavelength.
Louise
CD Track spacing
Posted: Tue Feb 20, 2007 8:00 pm
by GRAHAMRICHARDS
the laser pointer we are using was $15 cdn (inexpensive). on the laser pointer it listed a "range" of wavelengths so i took an average.
Re: CD Track spacing
Posted: Tue Feb 20, 2007 8:14 pm
by Louise
GRAHAMRICHARDS wrote:Again I'm looking for an anwer of 500 nm
thanks
calculate laser pointer wavelength as follows
630+680 = 655 nanometres
2
I checked the angles with my son again and still getting answer of 2-4 Um.
Here's the angles. Note didn't use m=0.
Incident angle 20 (70 on protractor)
+1 30 (60 on protractor)
+2 50 (40 " ")
-1 15 (75 " ")
-2 0 (90 " "}
Incident angle 30 (60 on protractor)
+1 57 (43 on protractor)
+2 75 (15 " ")
-1 30 (60 " ")
-2 10 (80 " "}
Incident angle 40 (50 on protractor)
-1 30 (60 " ")
-2 3 (87 " "}
Okay, something is odd here. First, I think in the first dataset, -2 should not be zero. This is pretty close to the setup in the picture, and they show a 30 degree angle.
In the second dataset (30 degrees) I think the diffraction orders are wrong. The beam coming out at 30 degrees is n=0, right? See, for example "The reflected ray (when θm = θi) has order 0 (zero)."
Lastly, the thread zzzzdoc pointed you to makes a very important point- the sign(+ or -) of the angle must be entered correctly. I got close to the correct answer for several of your sets of numbers (and having the sign right was key!), but there are definitely errors in some of the values.
Hope this helps!
Louise
CD Track spacing
Posted: Tue Feb 20, 2007 11:26 pm
by GRAHAMRICHARDS
It may be how I am reading the angles on the protractor and entering the angles into the equation.
90 on the protractor is entered as 0
to the left of 0/90 + or - ?
to the right of 0/90 + or - ?
I thougth the +/- was relative to the reflected angle of the incident beam.
thank you for assistance.
getting the sign right
Posted: Thu Feb 22, 2007 7:55 pm
by scibudadmin
Graham,
Sorry for your troubles with this project.
I had to refresh my memory on this project. Getting the sign right is a bit tricky, and the instructions don't help enough. The easiest way is to show you an example.
Take a look at the example photograph in the project (here's the link:
https://www.sciencebuddies.org/mentorin ... p011.shtml and the photo is in step 2 in the Experimental Procedure).
The incident angle is 20 degrees.
The diffracted beam for order m=+1 is at about 48 degrees. It's on the opposite side of normal from the incident angle, so it is positive. Calculating d we get: 1 x 655 (sin 48 - sin 20) = 1,633 nm = 1.6 um
The diffracted beam for order m=-1 is at about 7 degrees. It is on the same side of normal as the incident beam, so it gets a negative sign. Calculating d we get: -1 x 655 (sin (-7) - sin 20) = 1,412 nm = 1.4 um.
The diffracted beam for order m=-2 is at about 34 degrees. Again, since it is on the same side of normal as the incident beam, it gets a negative sign. Calculating d we get: -2 x 655 (sin (-34) - sin 20) = 1,453 um = 1.4 um.
Make sure that you aren't mistaking the reflected angle (m=0) for one of the diffracted beams. I'll edit the project to make it clear how to determine the sign.
Best regards,
Andrew Olson, Ph.D.
Senior Scientist,
Science Buddies