My Report on Experiment & Error Analysis

[bayan]

This is my report on an experiment that I have done and I really needed the latex codes to generate the equations. I have done the other parts of the report on other PC.

Do they seem to be ok? (the errors, like the max possible value and min possible value)

no help is really needed, rather I want to know what you think of it, any where I can improve or anything I should drop (unnecesary things)

Thank you!

for \lambda
\lambda =\frac{18.5}{6.5}
\lambda = 2.85 cm


This indicates that the apprxmiate wavelength is 2.85 cm with a error margin of ^+_-0.03

now to find the spacing between the ball bearings we can use the first angle and Bragg's equation to find the other angles.

n1

1\lambda =2dSin14^o

d=\frac {2.85}{2Sin14^o}

d= 5.89 cm


Now we can find values for other angles that correspond to this equation.

n2

\theta=Sin^-^1 \frac {5.7}{11.78}

\theta = 28.9^o

for n3

\theta= Sin^-^1 \frac {8.55}{11.78}

\theta = 46.5^o

for n4

\theta = Sin^-^1 \frac {11.4}{11.78}

\theta = 75.4^o

there is no n5 as the equation makes no scence and there is no value for sin grater than 1





Now we come to the errors that were involved in this experiment!

The wavelength was within a error range of ^+_-0.03 cm.

the angle was read to the nearest degree so it was really 14^o ^+_-1^o which then makes the d value lower and higher.

For example the maximum d is when angle was 14^o-1^o=13^o

Now we get the new d=\frac {2.88}{2sin13^o} which is 6.4 cm

The minimum value obtainable from these sets of result are as followed.

d=\frac {2.82}{2sin14^o} which is 5.82 cm

For n2 the angle was 28.9^o which could have been altered such that maximum is \theta = sin^-^1 \frac {5.76}{11.64} which is about 29.7^o and the minimum value could have been
\theta = Sin^-^1 \frac {5.64}{12.8} in which the angle is about 26.1^o

This just repeats until the last value of reflection angle.

I will show that the last angle would still exist even with the big error probability.

For Max \theta the \theta=2.88 which results \theta = sin^-^1 \frac {11.52}{11.64} which is 81.8^o and for Min the angle is \theta = sin^-^1 \frac {11.28}{12.8} which happens to be about 61.8^o

So infact the errors did not have a huge impact on the resultst that we were really interested (which is to find how many ball bearings there is!) but if we were to have a diffrent aim those errors would have made it really hard (for example if we wanted to see what exactly the crystall looks like)

 

[Moonbear]

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