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Some PHY4 things to clear up.
I would appreciate if someone could help me with the following questions:
1. What is the difference between the phase difference and path difference? How do you calculate each of them? - I remember seeing a question to calculate them in radians but I don't know how that works.
2. The photoelectric emission on circuits has the emmiter as positive or negative, to which the monochromatic radiations points to? Is this the cathode or anode? - I keep getting confused as many past papers use differently; one used the positive terminal as cathode and negative terminal as anode. A diagram would help me.
Thank you.
1. What is the difference between the phase difference and path difference? How do you calculate each of them? - I remember seeing a question to calculate them in radians but I don't know how that works.
2. The photoelectric emission on circuits has the emmiter as positive or negative, to which the monochromatic radiations points to? Is this the cathode or anode? - I keep getting confused as many past papers use differently; one used the positive terminal as cathode and negative terminal as anode. A diagram would help me.
Thank you.
1. Phase difference/angle - the stage the simple harmonic motion has reached, hence it could be from 0 to 2pi. (0/2pi signifies constructive interference, completely in phase, while pi signifies destructive interference, completely out of phase) Path difference is simply How much further one wave is relative to another. The path difference two waves could be very large, but their phase difference could be 0, since shm is periodic.
2. Photoelectric cell/emitter is the source of electrons, so is the cathode (analogous to cathode in cathode ray oscilloscope). It "points to" the anode.
http://www-personal.umd.umich.edu/~devlin/phys305/Krane02.gif
The important thing is you know the direction of electrons, and the positive terminal is called the anode. The other is the cathode.
2. Photoelectric cell/emitter is the source of electrons, so is the cathode (analogous to cathode in cathode ray oscilloscope). It "points to" the anode.
http://www-personal.umd.umich.edu/~devlin/phys305/Krane02.gif
The important thing is you know the direction of electrons, and the positive terminal is called the anode. The other is the cathode.
Eau
1. Phase difference/angle - the stage the simple harmonic motion has reached, hence it could be from 0 to 2pi. (0/2pi signifies constructive interference, completely in phase, while pi signifies destructive interference, completely out of phase) Path difference is simply How much further one wave is relative to another. The path difference two waves could be very large, but their phase difference could be 0, since shm is periodic.
2. Photoelectric cell/emitter is the source of electrons, so is the cathode (analogous to cathode in cathode ray oscilloscope). It "points to" the anode.
http://www-personal.umd.umich.edu/~devlin/phys305/Krane02.gif
The important thing is you know the direction of electrons, and the positive terminal is called the anode. The other is the cathode.
2. Photoelectric cell/emitter is the source of electrons, so is the cathode (analogous to cathode in cathode ray oscilloscope). It "points to" the anode.
http://www-personal.umd.umich.edu/~devlin/phys305/Krane02.gif
The important thing is you know the direction of electrons, and the positive terminal is called the anode. The other is the cathode.
I am wrong, in that I said phase difference is from 0 to 2pi. Apparently according to markschemes, phase difference could could go way beyond 2pi, e.g. 3pi, 100000000pi etc.
So for phase difference: even pi (starting from and including zero) means constructive interference, odd pi (starting from pi) means destructive interference
For path difference: integeral multiple of wavelength (starting from and including 0) means constructive interference, and half-integeral multiples of wavelength (starting from 1/2 wavelength) means destructive interference.
Thanks for that Eau, I also found out about a formula which is: phase difference=[2pi/(lambda)]*path difference.
According to the equation for centripetal force, F=mv^2/r, on these types of roads, the velocity cars travel is very large. In order for cars to turn safely around bends at high speeds, the radius of their curved path must be big (r must be large) in order for the centripetal force to be easily provided by the friction. If the radius of the path was small, i.e. sharp bends, then the centripetal force needed for the car to maintain and go around a curved path would be too large to be provided by the friction, hence the car would leave the highway.
1misty
cud ne 1 explain y roads designed 4 high speed hav no sharp bends
According to the equation for centripetal force, F=mv^2/r, on these types of roads, the velocity cars travel is very large. In order for cars to turn safely around bends at high speeds, the radius of their curved path must be big (r must be large) in order for the centripetal force to be easily provided by the friction. If the radius of the path was small, i.e. sharp bends, then the centripetal force needed for the car to maintain and go around a curved path would be too large to be provided by the friction, hence the car would leave the highway.
1misty
cud ne 1 explain y roads designed 4 high speed hav no sharp bends
Circular motion in sharp bends, where F=mv2/r.
High speeds, mean you need a very large resultant force (provided by friction) to maintain the circular motion at that "radius". However, there is a limit to how much the friction could provide, and well you will fly off (to a larger radius to compensate for the limited force provided).
ma3eeni
Does anyone know where I can find a site that describes the double slit experiment using videos or stuff. I am so confused with that. Specially the single slit diffraction effect.
You don't need to understand the effect, just memorise it: symmetrical fringes, central maxima is double the width and brighter.
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