Young's Double Slit Experiment
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- Опубликовано: 12 июл 2024
- This physics video tutorial provides a basic introduction into young's double slit experiment. It explains how to calculate the distance between the slits given the wavelength of light used and distance from the slit to the screens. It also explains how to calculate the distance between the first order bright fringe and the central bright fringe. The bright fringes are the locations where constructive interference occurs and the dark fringes are produced due to destructive interference. This tutorial contains many examples and practice problems.
Ray Diagrams:
• Ray Diagrams
The Huygens Principle:
• Huygens Principle - Ph...
Young's Double Slit Experiment:
• Young's Double Slit Ex...
Single Slit Diffraction:
• Single Slit Diffractio...
Diffraction Grating Problems:
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Polarization of Light Problems:
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Planck's Constant and Blackbody Radiation:
• Planck's Constant and ...
Photon Momentum and Effective Mass:
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Wien's Law:
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Compton Effect & Wavelength:
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Physics 2 Final Exam Review:
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Hey, on question 2, I got the answer as 3.3*10^-11 meters, could there have been a conversion error on your part because I converted all the values into meters via calculator converter.
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12:45
Multiply 8.5 x 3
Multiply with wavelength
Divide number with 0.5
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Look at the RUclips video "Double Slit Experiment" by Todd Grigsby.
It uses an interferometer to show that light waves can bounce off another.
This applies to the double slit experiment thus the title.
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Thank you sir for this veryy understanding video! But you only teach about the bright fringe and i want to know about dark fringe. Im having problem with determining the m of dark fringe.
for dark fringe, change lambda to lambda/2 in the equations
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why is theta equivalent to theta 1 at 4:00 minutes into the video? I wanted to know why those 2 angles are equivalent. tks
If the distance between the two points is 6cm and the angle of reflection, is 30⁰ what is the wavelength?
we just did this in class
Is sine-product which is the dark of the pupil sight
anwer of the third question : 1.76*10^-3 m/1.33 = 1.32*10^-3
right?! that is what i got as well, i am unsure why he isnt dividing by the m (canceling m)
On the last example, why didn't we multiple m=4 on the left side of the equation?
u waffled so hard
what about darker sides? do we just add 1/2lamda ?
for the final example, when calculating for d, why are you not dividing by m? I see you are leaving it on the numerator side of the equation when dividing by the remaining left side, but not including it on the denominator.... which would cancel m out entirely... maybe I am missing something.. someone explain pls
for 10:35 instead of using dy(1)=lambda m D you could've just used tan1.5 times 4.5 (make your life easier)
How to consider angle is small?
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What would happen if the constructive interferencing from 1 set of double slits passed through another set of double slits? Would the universe break because of energy being created?
😂😂😂 it's very likely that's you'd not be able to align the double slits at the exact same spot where the waves are in phase, but if you do... You'll just get another interference pattern where the intensity/amplitude is lesser than the last interference pattern you saw.
The highest intensity in an interference pattern is achieved at the centre of the double slits (projected on the screen) but if you plan on using another 'Double' slit, the starting intensity of the light would be less than the original, and hence you'd only get a fainter interference pattern :)
If we are asked to calculate lamda or width fringe of first minima are we going to use m=0
Or m=1
Helppppp
I believe you have to use one of these 2
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i don't get why the two angles (theta 1) are equivalent. could you help me to prove it please?
thank you
It has to do with the geometry of similar triangles
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I learned this last summer in CSA at CalTech.
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A wave is spread across. In a double slit experiment, if the wavefunction representing the electron, hits the slits, shouldn’t that be a measurement and shouldn’t that collapse the wavefunction? Now, you'll say it's not a physical wave, it's a probability wave. But then how does a probability wave split into two after the slits? Then, it should be like, the wave hits the slits, the electron says, dude I'm going through the slits, so don't collapse, but you can split into two and diffract. When does it decide that it's going through the slit?
Why did they divide the distance of the two slits by 10^3??
Can we use white light in this expriment ? If yes how ? If not why ?
No, because white light is a mixture red, blur, green, ect., which are at different wavelengths. This experiment uses light at just one wavelength. If you use light with different wavelengths, then each wavelength is going to be diffracting at different angles and interfering with each other so you won't get the neat maximum-minimum pattern on the screen
Yes, you can use white light, it will just produce an even cooler looking interference pattern as the different colours will diffract at a different angle, leaving you with something similar to a horizontal looking rainbow
Monochromatic light is preferred because the resulting interference pattern is more sharper when compared to a white light (which is a mixture of other colours).
White would give you interference, but not as sharp and distinct as a monochromatic source.