To have a good understanding on polarisation, you will need to know how polarisation works. There is actually a distinction between mechanical or EM waves in relation to polarisation.
For mechanical waves, you can think of a transverse wave generated by a rope. When the slit is in the same plane of oscillation of the wave, the wave will pass through, and when it is an a certain angle, only a component of the wave will pass through, hence being polarised.
However, for EM waves, which are oscillating E and B fields which are mutually perpendicular to each other, the polarisation is due to energy absorption by the materials. A polaroid consists of long chains of hydrocarbon that are aligned in a particular direction. The valence electrons of the molecules can move easily along the chains. Only under the application of the electric field the electrons will then move and hence absorbing energy from the field. Hence when the plane of oscillation of the electric field either in the same direction or has a component as the hydrocarbon direction, then the energy in that direction will be absorbed. In other words, the energy in the perpendicular direction (which is called the polarising axis) will not be absorbed and hence passed through. When the E-field is reduced due to polarisation, the B-field will also be reduced proportionally.
Saturday, September 09, 2006
Thursday, September 07, 2006
Adam asked:Why does the B-field not get cancelled out during polarisation?
I supposed you are trying to ask about polarisation of EM waves--- which is basically an oscillating B and E field that are mutally perpendicular to each other. If the polarising axis is parallel to the E-field, then the E-field gets out without any attenuation. So the E-field will oscillate with the same amplitude. Any E-field that is present will always be accompanied by the B-field which is always perpendicular to it, so the B-field will also remain unattenuated. If the polarising axis is at an angle to the B-field, then we have the situation of the E-field being rotated by the same angle and the magnitude will just be the component parallel to the E-field. Once E-field is reduced, the perpendicular B-field (which has also rotated to be always perpendicular to the E-field) will also have a smaller amplitude.
Will give u more info another day when i have more time to write....
Will give u more info another day when i have more time to write....
Monday, September 04, 2006
More on photoelectric effect: Will frequency affect current?
My last post said, "the simple answer is...."
So what's the more "complex" answer? Dunid to worry abt tis more complex answer if u r still not too familiar with the more basic concepts... so here goes...
In fact, freqency can affect the photoelectric current.... but why?
Each electron on the metal requires a different amt of energy (from a min value called to work function to extremely high value) to release it, so not every photon can release an electron, it is a probability thing. In fact, for every electron to be released, many many photons might be required, and exactly how many will depend on factors such as the type of metal involved, the frequency of the EM radiation, and even physical conditions such as temperature.
An electron can only be released from the metal if it so happens that the photon that hits a particular electron has energy at least equal to that required to release it. With higher frequency, the energy of each photon would be higher, so the chance that the electrons can meet with a photon with sufficient energy to release it would be higher, so statistically there will be a greater number of photoelectrons released per unit time. If that is the case, the current would also be higher.
Note: the frequency range might not be very large to give rise to a very significant change in the photoelectric current. in other words, the photoelectric current might not be affected much in practice.
So what's the more "complex" answer? Dunid to worry abt tis more complex answer if u r still not too familiar with the more basic concepts... so here goes...
In fact, freqency can affect the photoelectric current.... but why?
Each electron on the metal requires a different amt of energy (from a min value called to work function to extremely high value) to release it, so not every photon can release an electron, it is a probability thing. In fact, for every electron to be released, many many photons might be required, and exactly how many will depend on factors such as the type of metal involved, the frequency of the EM radiation, and even physical conditions such as temperature.
An electron can only be released from the metal if it so happens that the photon that hits a particular electron has energy at least equal to that required to release it. With higher frequency, the energy of each photon would be higher, so the chance that the electrons can meet with a photon with sufficient energy to release it would be higher, so statistically there will be a greater number of photoelectrons released per unit time. If that is the case, the current would also be higher.
Note: the frequency range might not be very large to give rise to a very significant change in the photoelectric current. in other words, the photoelectric current might not be affected much in practice.
Photoelectric effect: Will frequency affect current?
A group of students recently asked mi tis question... how will the frequency affect the photoelectric current?
The simple answer is frequency will NOT affect the photoelectric current, and the standard response is this: frequency will only affect e energy of individual photon (since E=hf), but the number of photons that will incident on the metal surface per unit time will not change, hence the number of photoelectrons emitted will not be affected. This also means that the charge arriving at the collector's plate per unit charge will not change, implying the same current.
Some students will then argue, when the energy of the photon is higher, wouldn't the max KE be higher, hence the velocity of the photoelectrons reaching the collector's plate be higher, and thus higher current?
Well, it is perfectly correct to say that with higher photon energy, the max KE is higher and hence the velocity will also be higher. However, this does not mean that the current is higher, as the rate that the photoelectrons reaching the collector's plate, and hence the quantity of charge, will not be affected. Remember I = ne/t (e is a constant, n/t is the number of photoelectrons reaching the collector's plate per unit time)
I gave this analogy when I was explaining to my students... with slight modification here :)
If all e students in e audi want to go to the toilet, but I only allow one student to go every min. naturally the first couple of stds dun quite feel e urgency n take their own sweet time to go to the toilet, but towards the end the students would rush to the toilet! Do I have more students going to the toilet per min? Of cos not, 'cause i m still controlling the rate of students leaving! In photoelectric effect, this rate is proportional to the quantity known as intensity, and the energy of each student is like the energy of the photons, each of which is proportional to the frequency!
Note: Intensity is actually equal to the energy incident on the metal surface per unit time per unit surface area, which is also equal to (N/t)hf/S, where S is the surface area of the metal. However, when we say thatthe intensity is changed, by default it is assumed that the frequency remains the same (freq characterises the type of EM waves, ie the 'colour'), and hence by changing the intensity would mean changing the rate of incident of photons on the metal.
The simple answer is frequency will NOT affect the photoelectric current, and the standard response is this: frequency will only affect e energy of individual photon (since E=hf), but the number of photons that will incident on the metal surface per unit time will not change, hence the number of photoelectrons emitted will not be affected. This also means that the charge arriving at the collector's plate per unit charge will not change, implying the same current.
Some students will then argue, when the energy of the photon is higher, wouldn't the max KE be higher, hence the velocity of the photoelectrons reaching the collector's plate be higher, and thus higher current?
Well, it is perfectly correct to say that with higher photon energy, the max KE is higher and hence the velocity will also be higher. However, this does not mean that the current is higher, as the rate that the photoelectrons reaching the collector's plate, and hence the quantity of charge, will not be affected. Remember I = ne/t (e is a constant, n/t is the number of photoelectrons reaching the collector's plate per unit time)
I gave this analogy when I was explaining to my students... with slight modification here :)
If all e students in e audi want to go to the toilet, but I only allow one student to go every min. naturally the first couple of stds dun quite feel e urgency n take their own sweet time to go to the toilet, but towards the end the students would rush to the toilet! Do I have more students going to the toilet per min? Of cos not, 'cause i m still controlling the rate of students leaving! In photoelectric effect, this rate is proportional to the quantity known as intensity, and the energy of each student is like the energy of the photons, each of which is proportional to the frequency!
Note: Intensity is actually equal to the energy incident on the metal surface per unit time per unit surface area, which is also equal to (N/t)hf/S, where S is the surface area of the metal. However, when we say thatthe intensity is changed, by default it is assumed that the frequency remains the same (freq characterises the type of EM waves, ie the 'colour'), and hence by changing the intensity would mean changing the rate of incident of photons on the metal.
why tis Physics blog?
every now n then i would ve some inspirations regarding physics ideas (although i mus confess tat i dun tink abt physics realli tat much ;). Especially after stds asked mi very interesting n sometimes thought provoking questions....
tink i should share wif all so tat everyone will benefit :)
tink i should share wif all so tat everyone will benefit :)
First Post for Smart Guide to Physics
I created another blog for the Physics freaks! Wanted to name tis blog "Idiot Guide to Physics", inspired by my 05S7A student vanessa, but no one in Hwa Chong is an idiot rite? haha... shall call it... "Smart Guide to Physics" lah :)
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