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.(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]\lambda[/tex]

[tex]\lambda =\frac{18.5}{6.5}[/tex]

[tex]\lambda = 2.85 cm[/tex]

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

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

[tex]n1[/tex]

[tex]1\lambda =2dSin14^o[/tex]

[tex]d=\frac {2.85}{2Sin14^o}[/tex]

[tex]d= 5.89 cm[/tex]

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

[tex] n2 [/tex]

[tex] \theta=Sin^{-1} \frac {5.7}{11.78}[/tex]

[tex] \theta = 28.9^o [/tex]

for [tex] n3 [/tex]

[tex] \theta= Sin^{-1} \frac {8.55}{11.78} [/tex]

[tex] \theta = 46.5^o [/tex]

for [tex] n4 [/tex]

[tex] \theta = Sin^-1 \frac {11.4}{11.78}[/tex]

[tex]\theta = 75.4^o[/tex]

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

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

The wavelength was within a error range of [tex]^+_-0.03 cm[/tex].

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

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

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

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

[tex]d=\frac {2.82}{2sin14^o}[/tex] which is [tex]5.82 cm[/tex]

For [tex]n2[/tex] the angle was [tex]28.9^o[/tex] which could have been altered such that maximum is [tex]\theta = sin^-^1 \frac {5.76}{11.64}[/tex] which is about [tex]29.7^o[/tex] and the minimum value could have been

[tex] \theta = Sin^-^1 \frac {5.64}{12.8}[/tex] in which the angle is about [tex]26.1^o[/tex]

This just repeats untill the last value of reflection angle.

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

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

So infact the errors did not have a huge impact on the resultst that we were really intrested (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)

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# Homework Help: My Report

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