I had a question: When both paths are equal, why is it bright in Det Y and dim in Det X specifically? Why can't it be the opposite? Does it have to do with the order of reflection-transmission of the beam component? I mean, for Det Y measurement, path x is transmission + reflection, while path y is reflection + transmission (both paths have reflection as well as transmission...is this why interference is constructive?). In case of Det X measurement, path x is transmission + transmission while path y is reflection + reflection (Does this lead to destructive interference?). If my assumption is correct, then sending the photon vertically from top to BS1 (instead of horizontally like here) would lead to Det X being bright while Det Y being dim?
This is correct. The particles get a phase of +Pi depending on the traveled path. When the traveled distance is the same, you will only see particles at Det Y because the phase shift leads to destructive interference at Det X. Sending photons from top to BS1 will lead to the opposite outcome: all particles will hit Det X
They never produce single photons, they are producing single separate individual wavelengths, like this ~ ~ ~ ~ ~. Photons do not exist. The photoelectric effect is easily explained by blue light having the right frequency of pos/neg magnetic induction to magnetically agitate a negatively charged election out of it's shell orbit. Electromagnetic Radiation (EMR) is waves at all frequencies, all the experiments show light is waves, there is no experiment which shows light is 'particles'.
Excellent explanation, thank you!
Best video on this theme ever! Thank you
Really nice video
Thank for uploading!!
I had a question: When both paths are equal, why is it bright in Det Y and dim in Det X specifically? Why can't it be the opposite? Does it have to do with the order of reflection-transmission of the beam component? I mean, for Det Y measurement, path x is transmission + reflection, while path y is reflection + transmission (both paths have reflection as well as transmission...is this why interference is constructive?). In case of Det X measurement, path x is transmission + transmission while path y is reflection + reflection (Does this lead to destructive interference?). If my assumption is correct, then sending the photon vertically from top to BS1 (instead of horizontally like here) would lead to Det X being bright while Det Y being dim?
This is correct. The particles get a phase of +Pi depending on the traveled path. When the traveled distance is the same, you will only see particles at Det Y because the phase shift leads to destructive interference at Det X. Sending photons from top to BS1 will lead to the opposite outcome: all particles will hit Det X
If mirrors are replaced with detectors in interferometer to send the photon back will we see interference pattern at the source?
Excellent! TU
Great video. Thanks!
They never produce single photons, they are producing single separate individual wavelengths, like this ~ ~ ~ ~ ~. Photons do not exist. The photoelectric effect is easily explained by blue light having the right frequency of pos/neg magnetic induction to magnetically agitate a negatively charged election out of it's shell orbit. Electromagnetic Radiation (EMR) is waves at all frequencies, all the experiments show light is waves, there is no experiment which shows light is 'particles'.
Well, publish a paper instead a YT comment if you are sure that 100 yrs of Physics are wrong