#1
I'm doing an experiment that involves the mechanical determination of the speed of light (through air). In short, a laser is shot through a lens (1) to focus it to a point image at s, then through another lens (2) a distance away. The beam then hits a rotating mirror and is reflected to a spherical mirror a distance D away. The beam then travels back to the rotating mirror, through lens (2), and through a beam splitter and microscope apparatus. The displacement between the returned image at point s' and the original image at point s can be used to determine the speed of light:

c=[4AwD^2]/[(D+B)ds']

Where:
c= speed of light
A= distance between lens (1) and lens (2) minus the focal length of lens (1) ~1mm
B= distance between lens (2) and the rotating mirror ~1mm
D= distance between the rotating and fixed, spherical mirror ~1mm
w= rotational velocity of the rotating mirror in rad/s ~1rev/s
ds'= the displacement of the image point ~.005nm

* ~ means measurements can be made to within the listed value.

As part of the experiment I need to derive an expression for the error in the speed of light calculation based on the error in each term. Can anyone help with this part?

If there is confusion, you can check out Foucalt's experiment - my experiment is a replication.
#2
Well usually when I make an error I express it as Awww Fuck!!!!!
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#3
Don't expect anyone to be doing physics as advanced as yours. Just a warning

OK, I read it. I just multiply the percentage errors. If that's what you're looking for
Last edited by Shred Head at Dec 8, 2008,
#4
I will read then edit my post but you will find a lot of "lol wut?" based answers in the pit
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#5
it's true... I mean, google has it all

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#6
what exactly do you mean with "expression for the error" (I got myself some university physics, which are in english but i've never heard of that one)
#8
give them your percent error.
using this fromula


absolute value of(known value-experimental value) divided by the known value, then multiply that number to get a percent
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wait.... wuttt???
#9
Basically you have to take the individual errors of each value you're measuring. The absolute error on a measured value is half the reading value of your equipment. Now if you add or substract 2 readings, you have to add the mistakes on them.
It can also happen that you have to multiply 2 read values. In this case you'll have to use the relative error.

The relative error is the absolute error divided by the number you measured. Make sure that the counter in your fraction is 1 (just an agreement). EG: if you measure with a common lat, you'll usually measure in cm (here in europe). This means that you could measure 5cm, with an absolute error of 0.5cm. Your relative error would be (0.5/5) = (1/10). Now if you divide or multiply 2 measured values, you can calculate the relative error on your outcome by adding the relative errors on them.

So if you'd multiply 5cm +-0.5cm with 10cm +-0.5cm, you'd get 50cm +-(the sum of the relative errors)
in this case, they are 1/5 and 1/10, which makes the relative mistake 1/(10/3) (= 1/3.33...)
Now that's the relative mistake so you'd have to multiply that by 50 again
in other words it's 50cm +-15cm (I don't have to tell you that that's huge, do I? xD).

I hope this helps, cause I didn't take the time to read the experiment entirely. What you should do is take the formula expressing c, and use the rules I just explained to calculate the total error on c by means of the errors of A, B, D, W and ds.

Good luck on that, I'm not gonna calculate it all for ya. It's dirty work...

EDIT: take that shread head xD
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haha

*wipes tear from eye*
Oh you're good.
Last edited by Base Ics at Dec 8, 2008,
#10
Additions and subtractions: Add error margins

Multiplications:

A= Value 1
B= Value 2
a= Error for A
b= Error for b
S= A*B
s=error margin of S

s = S [(a/A) + (b/B)

I can't remember what to do for divisions though...
#12
I did this the other day. It sucked. I still don't understand how it works.

But with the error, just find the generalised error formulae (in your work or on google) and then just stick them in. I was stuck on it for a bit but it's quite easy once you actually take a look.

Quote by Base Ics
So if you'd multiply 5cm +-0.5cm with 10cm +-0.5cm, you'd get 50cm +-(the sum of the relative errors)
in this case, they are 1/5 and 1/10, which makes the relative mistake 1/(10/3) (= 1/3.33...)
Now that's the relative mistake so you'd have to multiply that by 50 again
in other words it's 50cm +-15cm (I don't have to tell you that that's huge, do I? xD).


You square the fractional errors when you multiply (or divide) two thing.

(dF/F)^2 = (dA/A)^2 + (dB/B)^2

Where F (or A or B) is the value and dF (or dA or dB) is the error in F, A or B. Using that formula and the one for addition/subtraction:

dF = dA + dB

You can combine them all to figure out a formula for the error in c.
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#13
Is this basically the Michelson-Morely interferometer experiment? Either way you should have been supplied with a lab book at the beginning of the year detailing, in full, how to determine your errors (compound, observational and possibly random).

If they didn't then there's a lot of great resources on the web. I can't imagine why you'd choose to ask UG when there's physics forums out there

Google "hyperphysics"; that site normally helps me out when I'm trying to grasp something new.
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#14
I think the formula you may be looking for is something like this :



Where the delta signs are the errors in the values. For every value that you have an error, you just add another one of the (error/|value|)^2

I think this is called adding errors in quadrature..