Episode #125 of the Stack Overflow podcast is here. We talk Tilde Club and mechanical keyboards. Listen now
3

This is not an incandescent bulb. It's an LED bulb with strings of LEDs on wires arranged to look like incandescent filaments. These LEDs are blinking too fast for the eye to see, but the camera's exposure is fast enough to record moments when one or the other "filament" is unlit, leading to the appearance of flashing as the LED flashing frequency beats ...


2

With monochromatic illumination of a double-slit interferometer, there is no spectral modulation. The resulting fringe pattern is monochromatic. With white light illumination, the resulting fringe pattern is a superposition of monochromatic fringe patterns, each with fringe spacing that corresponds to its wavelength. So, the resulting fringe pattern has ...


2

Your question is an excellent one, and puzzled me when I was studying for my PhD in quantum theory. The answer is that the intermediate screen in which the slits are formed can indeed collapse the wave functions of electrons projected towards it; but they are the electrons that make it through neither slit, being instead absorbed or scattered by the ...


1

I've heard it said (a while ago) that we don't usually speak of a single photon being linearly polarized or unpolarized, but rather we can speak of an ensemble of photons that are linearly polarized or unpolarized. Is this still the current thinking in Quantum Optics? No. It is perfectly possible to assign polarization states to single photons. There ...


1

On the assumption that the screen is far away from the two slits and the angles involved are small the fringe separation is $\Delta x = \frac {\lambda D}{d}$ where $\lambda$ is the wavelength, $d$is the slit separation and $D$ the two slit to screen distance. So $\Delta x \propto D$. A diagram with the angles exaggerated is shown below with the radial ...


1

The exact path difference is: $$ \Delta = \sqrt{D^2 + \big(\frac w 2\big)^2} -D $$ $$ \Delta = D\big[ \sqrt{1 + \big(\frac w {2D}\big)^2} - 1\big] $$ $$ \Delta = D\big[ \sqrt{1 + \epsilon^2} - 1\big]$$ with $$ \epsilon \equiv \frac w {2D}$$ as a small parameter under $w/D \ll 1$. Now Taylor expand around $\epsilon = 0$: $$\Delta = D\big[ 1+\frac 1 2 ...


Only top voted, non community-wiki answers of a minimum length are eligible