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Great demonstration and explanation of the Young's and newton's rings experiment.
~Thanks a ton. :) :)
shubhamgore 1 week ago
omg thank you
theboywho2 1 month ago
where's the compensatory plate to make the path difference same in case we use polychromatic source?? here it makes no difference rather!!!
rockeemanable 5 months ago
Comment removed
rockeemanable 5 months ago
so nice n very interesting i like it n ful of info thantks......
sanam9640 5 months ago
LOL! I recognized that that was 3D Studio just by sight. I think I improved all my engineeirng presentations by a letter grade just by making cheesy graphics & animations (not that I'm disparaging what you've done here).
PhotosByStephan 10 months ago
@PhotosByStephan Haha, yeah, I started making animated videos for school reports way back in high school using my Amiga. It was a sure-fire way to get a good grade. Thanks for commenting!
etlund 10 months ago
Elitzur–Vaidman bomb-tester
sites.google.com/site/physicsphilosophyscience/physics/interferometer-a-bomb-tester
Stupefience 1 year ago
gud but the design portion is very less...
TheAppu1 1 year ago
Great shit! I'll use it on my physics final tomorrow!
jeremy251 1 year ago
thankyou so so much for this.. got an exam on it tumrw n it totally mkes sense now :)
desiflychick100 1 year ago
NEW THEORY OF TIME
angper 1 year ago
BRAVO !!!
romanistANIA 2 years ago
This has been flagged as spam show
L/4 because the ray run through it's path twice: so for a real shift of L/2 (that means from a peak to a belly of the wave) you need to move the object L/4. If the two ray are in phase, that means that the object have been moved with a distance equal to lambda L: because you move from a peak to another peak of the wave (in this case L = 2*Pi). So to perform a L/2 change (from a peak to a belly) you need to shift L/4: in fact L/4 * 2 = L/2. With lambda L i assume that is wavelenght.
carly81 2 years ago
Comment removed
carly81 2 years ago
good work, nice effort put up in 1995 :)
gill1447 3 years ago 2
I don't know about reconstructing the image, but it seems like a good idea. This vid was from my undergrad days and is about 13 years old so I just don't remember the details.
Thanks for the comments!
etlund 3 years ago
well done
one question
you said that you can measure displacement by seeing how the fringes swap colors, corresponding to a lamda/4 change in distance.
do you know whether an actual image of the sample can be reconstructed using the interferometer ... cause thats the basis for OCT imaging, which I wanna learn more about
sdwipster 3 years ago
Very informative and well-done.
Understandable even to those with a nontechnical background.
Many thanks!
Sivfpidrn 3 years ago
when the distances of the mirrors to the beamsplitter are the same, there should be no fringes anymore, something you didn't mention in your video. nice work!
bmgf300z 4 years ago
Great, but the little diagram of the Michelson with the symbol for 'viewer' being represented by the eyeball caused me to shudder !!!!
If I'm drawing laser-diagrams, I NEVER use the little eye symbol to represent viewer except in the case where the eyeball really should be at that location.
RoadRunnerLaser 4 years ago