# Homework Help: My Report

1. Oct 6, 2005

### 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 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 $\theta$ 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 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)

Last edited by a moderator: Oct 6, 2005
2. Oct 17, 2005

### Tom Mattson

Staff Emeritus
Sorry for the late reply. This thread got lost in the hustle and bustle of our recent upgrade. In case you're still interested, here's what I think of your report.

That second line needs units.

The error margin needs units.

If you want LaTeX to line up with regular HTML text, then you should use the itex tags (inline LaTeX).

In the first line it looks like it should be $n_1$, but I'm not sure because it seems to come from nowhere. Also, $sin$ should not be capitalized. Similar comments for $n_2$ through $n_5$.