# Homework Help: Question concerning experiment on Heisenberg's uncertanity principle

1. Mar 17, 2014

### MathiasArendru

1. The problem statement, all variables and given/known data
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 theres an equation that i don't understand, ill try and explain as well as i can as i cant use images.

So theres 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 isnt right is it? So i cant figure out how this works out, i've looked up alot 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

Last edited: Mar 17, 2014
2. Mar 17, 2014

### 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.

3. Mar 18, 2014

### 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?

4. Mar 18, 2014

### 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!

5. Mar 19, 2014

### MathiasArendru

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

Take care