What is Error Bar Calculation for Interference & Diffraction Experiments?

[agentas]

I have done experiment on interference and difraction. The principe of experiment was that:laser beam was going through some slit and interference pattern has been shown on the wall.I had to find the width of slit.I have done that by simply measuring the distance from slit to the wall and the distance from center of interference pattern to first maxima and putting measured data to equation d=(2n+1/2)liambda/sin(teta) I have found the width of the slit.But now the teacher is asking me to put error bars on this value of d.How I should find this error bar value.Please help

 

[mgb_phys]

How did you measure the position of the fringes? What accuracy did you measure this to? And so how uncertain was your angle value?

 

[Marshall10488]

an error bar shows your uncertainty on a graph. uncertainty is +/-0.5 of a scale devision to which you measured (put very simply) unless your measurements have ben used in calculations to get your points on your graph then you need to add and multiply your errors depending on what function they have performed.
the error bars look like a cross hair (kind of) a |-| and a vertical one at everypoint, showing the range that the point will fall under.
hope this helps

 

[arunma]

If you assume that the errors are quite a bit smaller than actual values of the measurements (which is usually a good approximation), then you can use the error propagation formula. If you have some quantity $z = f(x,y)$, then the error is,

$$\sigma_z = \sqrt{\left(\dfrac{\partial z}{\partial x}\right)^2 \sigma_x^2+\left(\dfrac{\partial z}{\partial y}\right)^2 \sigma_y^2}$$

However, as I learned (the hard way) in my first summer of graduate research, you have to be careful with this trick. It's based on taking the first term in the Taylor series of the function, and this isn't always a good approximation. For example, with the logarithm function, which we often use when plotting data in astrophysics, the first term in the Taylor expansion is only good for small values of the argument. But for the equation which gives the path difference between the light rays from two slits, I think you shouldn't run into any problems.

 

[agentas]

The angle was calculated by equation tan(teta)=y/L.where y distance from center of interference pattern to first maxima and L the distance from slit to the wall .The angle is 0.540 degrees.Talking about uncertanty: L was 3,710m y= 0,035m .So I think L could be 2mm smaller or bigger ,where y was very accurate.So how I can calculate those error bars?Please help

 

[agentas]

an error bar shows your uncertainty on a graph. uncertainty is +/-0.5 of a scale devision to which you measured (put very simply) unless your measurements have ben used in calculations to get your points on your graph then you need to add and multiply your errors depending on what function they have performed.
the error bars look like a cross hair (kind of) a |-| and a vertical one at everypoint, showing the range that the point will fall under.
hope this helps

I have found the value of d (width of the slit) to be 0,0000947m.So what is error bar for this value?How can I calculate ?.Please give me someone clear answer.I was using simple ruler for measurements.(5 meters ruler divided in mm)

 

[mgb_phys]

First estimate the error on distance to the screen (0.5mm?) and the error in the position of the fringe (1mm?)
Now we have to work out what error in the sin(theta) this would give.
The correct treatment is given by arunma but it's probably a bit complicated for you

You can think of sin() as A/B where A is the spacing and B is the distance to the screen, if you have an error in two numbers and multiply them together then the error in the result is the sum of the relative uncertainty in each.

So if you measure the screen distance as 1m with an error of 1mm the relative error is 1/1000 = 0.001 and the spacing is 100mm with an error of 0.5mm = 0.5/100 = 0.005
The total error is 0.006 (=0.001+0.005) or 0.6%
The error in the angle is then 0.6% so the size of your error bars in angle would be about this.

You can estimate the error in your result by seeing what range of lines you could draw through th epoints with error bars, or there are mathematical ways of telling you the error (see least-squares-fit)