So we have a light bulb emitting through a small slit. I read somewhere that a blackbody shining through a small slit acts like a perfect emitter, and I wanted to know why. I figured it must have something to do with Heisenberg's Uncertainty. Here's how I went about it:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]

The momentum of light is just E/c, so

[tex]p = \frac{E}{c} = \frac{h}{\lambda}[/tex]

and

[tex]\lambda^{2} = \frac{h^{2}c^{2}}{E^{2}} [/tex]

To find Δp, just take the derivative wrt λ

[tex]\frac{\Delta p}{\Delta \lambda}=\frac{-h}{\lambda^{2}}[/tex]

and

[tex]\Delta p = \frac{h}{\lambda^{2}}\Delta \lambda[/tex]

I ignore the negative because this is uncertainty in either direction. So..

[tex]\Delta x \Delta p = \Delta x\frac{h}{\lambda^{2}}\Delta\lambda[/tex]

Then I subbed back into the first eq and solved for ΔxΔλ

[tex]\Delta x \Delta \lambda \geq \frac{\lambda^{2}}{4\pi}[/tex]

thus

[tex]\Delta x \Delta \lambda \geq \frac{h^{2}c^{2}}{4\pi E^{2}}[/tex]

Is this why they say that the light shining through a small slit a near perfect emitter? Because since we don't know exactly where the light goes through the slit, it must have an uncertainty in the wavelength when it comes out. If so, what is that E for? Is it the energy of the photons before they enter the slit? Thanks.

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# Heisenberg's Uncertainty - ΔλΔx

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