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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 $\Delta x \Delta p_x \sim 8\pi\hbar$? |
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As others have remarked, Heisenberg's microscope represents nothing more than a scaling argument for the position-momentum uncertainty relation. This means the argument should be expected to lead to an expression that is accurate no better than within factors of order unity. Having said this, the optical (diffraction) limit for determining the blurring of a point source is well-defined and given by Abbe's criterion. See this Wikipedia page: http://en.wikipedia.org/wiki/Diffraction-limited_system . As you notice, the radius of the spot size is determined by the wavelength divided by twice the sine of the half-angle. You can determine the full width of the spot size by doubling this expression (it seems this is what you are hinting at?), if you do so, you end up at the equation used by the author you quote. |
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