Question concerning experiment on Heisenberg's uncertanity principle

[MathiasArendru]

Homework Statement

Hello guys my name is Mathias I'm 15 years and i have been tasked to show a little experiment in my school class. It's concerning Heisenbergs principle, you have probably all heard about this demonstration. It's basically firing a laser beam through a narrow passage, whereafter it will spread out due to σ x σ p ≥ h/4.
My problem is that there's an equation that i don't understand, ill try and explain as well as i can as i can't use images.

So there's not really any big need for me to explain the experiment or anything else, but just the equation, so here it is:

ΔPx=$\frac{h}{\lambda}$Sin($\Phi$)=$\frac{h}{\lambda}$Tan^-1($\frac{y}{x}$)

To me, this states that Sin($\Phi$)=Tan^-1($\frac{y}{x}$). But that isn't right is it? So i can't figure out how this works out, I've looked up a lot of trigonometry but nothing has helped me figure it out... Is it an error or is it me whos overlooking something?

If it's ok i have allowed myself to add a link to the document. It is in danish, but looking at the graphics on the first 2 pages, visualises my problem: http://kvucfysik.wikispaces.com/file/view/Lab+A+-+Kvantefysik.pdf
h=Plancks Constant

Thanks in advance, hope you can help me!
- Mathias

 

[BvU]

Hello mathias, and welcome to PF,

Yes. The pdf is a little inconsistent: b is the distance from the slit to the screen, a is the width of the peak, right?

On page 2 the figure on the top right has the letter b on the hypothenusa, but it would be better if they had put it on the lower rectangular side. Then you have $\tan \theta = { a\over 2} /b$.

Because a/2 << b you can write $\tan \theta = \theta$ without making a significant error.

 

[MathiasArendru]

Hmm... Yes b on the first figure page 1 is the length from the passage to the wall behind it. And confusingly on page 2 figure on the right, he refers b to the hypothenuse, however I'm fairly sure that in the equation I'm talking about, he's referring b to the hypothenuse.
Removing the irrelevant parts of the equation for this particular instance:
ΔPx=Sin($\Phi$)=Tan^-1($\frac{a}{2b}$)
But that can't be true? They don't relate that way, at least not when you ask my math book.
So how can that be?

I tried to make a imaginary triangle, Adjacent and Opposite sides being 5, the hypothenuse being 7.07. Obviously it has the angle opposite of hypothenuse as 90°, and the two other sides being 45°.
If i take the Sine of angle 45, i get 0,707, If i take the Inverse Tangent of the sides $\frac{5}{5}$ i get 45, as expected. So how can he relate these two? Did he maybe mean to take the Sine of the inverse tangent:?
Sin(Tan^-1($\frac{a}{2b}$))

Also
$\frac{a}{2b}$=$\frac{a/2}{b}$ Right?

 

[BvU]

Yes, you are right. $\tan \theta = { a\over 2} /b = {a\over 2b}$.
And ${\delta p_x\over p} = \sin\theta = \sin(\tan^{-1} {a\over 2b})$

Now, since the angles are so small, the discrepancies you notice are really minimal.
b is of the order of one or two meters, a is like 1 or 2 cm.
And tan^-1(0.01) = 0.0099997 only 0.0033 percent less than 0.01
Likewise, sin(0.01) = 0.0099995, only 0.005 percent less than 0.01

Even if it a/2b is as big as 0.1, the error is less than 0.5% !

You will never ever be able to measure with a better accuracy, so the error you introduce by letting ${\delta p_x\over p} = {a\over 2b}$ can be ignored completely!

 

[MathiasArendru]

I seem to have solved it now, thanks a lot for your consistent help BvU!

Take care