# What's the proper way to approximate the position uncertainty of a particle?

In this problem:

shouldn't $\Delta x\sim\lambda/\sin\theta$ be $$\Delta x\sim \frac{\lambda}{\sin\theta} - \left(\frac{-\lambda}{\sin\theta}\right) = 2\frac{\lambda}{\sin\theta}$$ instead such that the final answer is $\Delta x \Delta p_x \sim 8\pi\hbar$?

-
That's kinda beside the point, since $\lambda/\sin\theta\sim2\lambda/\sin\theta$ are of similar orders of magnitude. You get a different coefficient, but the important thing is that $\Delta p_x\sim c/\lambda$ where $c$ is "some constant"... – Alex Nelson Nov 30 '12 at 5:55
Hi, Alex.Yes, I actually do get that it's about the product of uncertainties relating to some constant but, I just wanted to make sure that, given the drawing, it should have been Δx ~ 2λ/sinθ. Was your comment a confirmation that I am correct to say that Δx ~ 2λ/sinθ? – Deniz Nov 30 '12 at 15:15
Well, if you want to know a more precise approximation, there are note online, also available here... – Alex Nelson Nov 30 '12 at 16:29
Hi, again. I am aware that the specific math is Δx Δp_x >= ħ/2. All I am looking for is a confirmation from someone as to whether Δx ~ λ/sinθ should be Δx ~ 2λ/sinθ for this specific problem/image above (assuming I am right). Edit: I realize that, on the surface, my question seems trivial but it's important for me for reasons that I have trouble verbalizing. – Deniz Nov 30 '12 at 16:49