Concept: Polarization Filters11m
Hey guys, in this video we're going to talk about this important quality of electromagnetic waves called the polarisation and these objects called polarisation filters let's get to it. Now what the polarisation is is it's the angle of the oh sorry it's the polarisation of an electromagnetic wave is the angle of the electric fields relative to some origin edge and relatives to some 0 degree angle. What do we mean by that angle I have four figures here where each red double headed arrow represents an oscillating electro sorry an oscillating electric field that's coming right at us. So what we're looking at is there are some oscillating electric fields and we are looking at it head on. So it's oscillating up and down up and down hence the double headed arrow now in the first image over on the right by the way each blue vertical line represents that origin that 0 degree angle that we're measuring the polarisation from in the left most image we have clearly a 0 degree polarisation no sorry it is polarized because it is at a particular angle but that angle is 0 it lines up with our origin here we have some acute angle It doesn't really matter what it is it could be 30 degrees 45 degrees 60 degrees etc. It's some degree sorry it's some angle that is less than 90 degrees and the third image we have a right angle we have a 90 degree polarisation now the polarisation is always going to be between 0 and 90 degrees because if you continue rotating that polarisation sorry that electric field you might get one that looks like this don't draw these guys just listen for the explanation you might get one that looks like this this is still going to have an acute angle.I didn't want you to draw because I was just going to erase it now all the way to my right right here what we have is a whole bunch of electric fields that are all at different polarisation angles now truly to represent this image we would need an infinite number of electric fields all of which have every possible polarisation angle and I can keep drawing more and more of them etc. This combination of all of these multiple waves that all have different polarisation angles and it comprises all the polarisation angles this we refer to as unpolarized light. So unpolarized light is made up of a whole bunch of different electric field orientations that compose all 360 of a circle.
So now we want to talk about what happens when light passes through a polarisation filter and what a polarisation filter is is it's something that filters out all light except for lights of a particular polarisation. Now let's consider initially unpolarized light as in the image all the way above me this unpolarized light. Right passes through a polarisation filter and this polarisation is choosing a particular angle right it's choosing a particular polarisation and all of the light other than the light that is at that particular angle gets filtered out all that light is gone, so some amount of light from the initially unpolarized light comes through and now has the polarisation chosen by that polarisation filter. The question is how much light do you lose? Well for polarisation and polarisation filter we track it by intensity and when initially unpolarized light like this unpolarized passes through a polarisation filter regardless of the angle of that polarisation filter the intensity always drops by half. Now a little bit of notation, this right here was called Isub0 or Inot. That variable Inot does not represent unpolarized light it's always used for whatever light passes into the polarisation filter regardless of whether it's unpolarized light or light that is already polarized because in the next equation I'm going to use Inot again but it's going to be for initially polarized light I just want you guys to be aware of that notation because your professors are going to use it your book is going to use it so it's very important that you understand the notation clearly so now let's consider what happens whenever initially polarized light passes through a polarisation filter or a polarizer the intensity drops like so I becomes Inot or Inot becomes Inot time's cosine squared of phi. Now phi is what I would call the polarisation difference. So relative to whatever 0 we have whatever origin we've measured polarisation against if you have some initial polarisation of 20 degrees and the polarising filter is 45 degrees the polarising difference is the difference between those angles 20 and 45 the difference is 25 that's what phi would be. So automatically, if you have a polarising filter at 90 degrees to the initial polarisation angle that's always going to result in a complete blockage of light no light is going to pass through in that scenario.
Let's do a quick example, unpolarized light passes through a polarisation filter with an angle of 45 degrees relative to the vertical this newly polarized light passes through a second polarisation filter with an angle of 55 degrees relative to the vertical if the intensity of the unpolarised light was 100 watts per meter squared. What's the intensity of the light after passing through the second polarising filter? So just for notation sake I'm going to call Inot the unpolarized light it passes through the first polarisation filter and then I'll call it I1 the results of the first polarisation filter then it will pass through a second polarisation filter and I'll call it I2 just for notation sake just to keep everything clear. So initially the light is completely unpolarized so we know that when it passes through any polarisation filter its intensity is going to drop by half so the first polarisation filter. I1 equals one half Inot which is one half of a 100 watts per meter squared that's what it started at right which is simply 50 watts per meter squared. Very easy unpolarized light going through a polarisation filter always drop by half now the second polarisation filter. Now we have a polarising difference we have light that is polarized to 45 degrees relative to the vertical that is the polarisation of our light when it leaves the first polarising filter the second polarising filter is 55 degrees relative to the vertical we want to apply the equation I2 equals I1 cosine squared of phi. Remember that this variable right here which in our equation I hope Inot is simply the intensity that enters the polarising filter in this case we call it I1 so the question is what is our polarising difference well if our polarising filter sorry our initial polarising filter was 45 degrees to the vertical and our final polarising filter is 55 degrees to the vertical then our polarising difference is 55 degrees minus 45 degrees which is simply 10 degrees.Our initial polarisation entering the second polarisation filter was 50 watts per meter squared times cosine squared of 10 degrees and this whole thing equals 48.5 watts per meter squared. Pretty straightforward application of equations polarisation remember guys that phi is not the angle of the polarising filter it's the polarisation difference it's the angle between the initial polarisation of the light entering it and the polarising angle of the filter, ok guys that wraps up our discussion on polarisation and polarising filters. Thanks for watching.
Example: Initially Vertically Polarized Light4m
Hey guys, let's do an example on polarisation initially vertically polarized light carries an intensity of 0.55 watts per meter squared if you had an adjustable polarizer what angle relative to the vertical would you want to align it so that the intensity of the light exiting the polarizer is 0.2 watts per square meter. So we know that our light is initially polarized it's saying it's initially vertically polarized so if you have initially polarized light entering a polarisation filter we have to use the equation I equals Inot cosine squared of 5 if the light was initially unpolarized then passing through any polarisation filter would just drop the intensity by half but this is not the case, so you always have to keep track is your light initially polarized is your light initially unpolarized in this case it's initially polarized and it's passing through an unknown angle that angle of the polarisation filter we can adjust all we know is the input intensity and we want to adjust it to get a particular output intensity so we know that the initial intensity is 0.55 watts per meter squared and we want the output intensity to be 0.2 watts per meter squared so we want to find what angle gives us this so what I'm going do is I'm going to divide Inot over and deal with the square root. So that tells me cosine of phi equals the square root of I divided by Inot which is just the square root of 0.2 divided by 0.55 and that's about 0.6. So if I plug this into a calculator to find out what phi my polarisation difference is that tells me that my polarisation difference is 53 degrees. So how would you align your polarisation filter what angle relative to the vertical would you align at you would align it 53 degrees from the vertical.
Right and that's because your initial polarisation was 0 degrees relative to the vertical and phi is your polarisation difference let's say this wasn't initially vertically polarized what if this was initially polarized 10 degrees to the vertical what would you where would you put your polarisation filter well 53 plus 10 would be 63 degrees that would be one option technically you can also go in the other direction and 43 degrees to the left of the vertical would also be an option. So just remember phi is not necessarily the final answer phi is the polarisation difference the final answer in this case is 53 degrees from the vertical. We found that by finding phi obviously, alright guys thanks for watching.
Problem: In each of the following cases, initially vertically polarized light enters the polarizing apparatus with the same initial intensity. Which polarizing apparatuses will cause the light to exit with the largest intensity, 90° from its initial polarization?
a) A single polarizing filter, oriented 90° from the vertical
b) Two polarizing filters, the first 45° from the vertical and the second 90° from the vertical
c) Two polarizing filters, the first 60° from the vertical and the second 90° from the vertical
d) Two polarizing filters, the first 30° from the vertical and the second 90° from the vertical.7m
A beam of polarized light of intensity I 0 travels in the +z-direction and is incident on two linear polarizers (LP1 and LP2) and a quarter wave plate (QWP) as shown below, LP1 has a transmission axes (TA) parallel to the x-axis, the QWP has a fast axis oriented 45° with respect to the x-axis and LP2 has a TA oriented 25° with respect to the x-axis. The light in Regions I, II, and III has intensities I1, I2 and I3, respectively. Find I3/I2.
a, I3/I2 = 0.82
b. I3/I2 = 0.18
c. I3/I2 = 1.0
d. I3/I2 = 0.88
e. I3/I2 = 0.50
The transmitted light beam (after the final polarized) will
a) have a vertical polarization direction.
b) have a polarization dependent on the direction of the incident polarization direction.
c) be unpolarized.
Unpolarized light of intensity 2.2 kW/m 2 passes through a polarizing filter oriented at 40° from the vertical. After passing through the filter, what is the electric field amplitude of the light? The magnetic field amplitude?
Vertically polarized light passes through a polarization filter with an axis 32° off the vertical. If the initial intensity of the light is 100 W/m2, what is the electric field amplitude of the light after it passes through the filter?
Unpolarized light of 120 W/m2 intensity passes through two polarizing filters. First, it passes through a polarizing filter with its axis aligned vertically, then it passes through a polarizing filter with its axis aligned horizontally.
(a) What is the intensity of the light after it passed through the filters?
(b) If a third polarizing filter was placed between the original two, with its axis aligned at 45o from the vertical, what would the intensity of the light be after it passed through all three filters?
Unpolarized light with intensity I0 is incident to two polarizing filters. The axis of the first filter is oriented at 37.0° counterclockwise from the horizontal, as shown in the sketch. The axis of the second filter is vertical. If the intensity of the light after it has passed through both filters is 0.300 W/m2, what is the initial intensity I0 of the light?
Unpolarized light with intensity 4.00 W/m2 is incident to two polarizing filters. The axis of the first filter is vertical. What is the angle between the vertical direction and the axis of the second filter if the intensity of the light after it passes through the second filter is 1.25 W/m2?
Unpolarized light of initial intensity 12.0 W/m2 is passed through three polarizing filters. Viewed in the direction the light is traveling, the axis of the first polarizing filter is vertical, the axis of the second polarizing filter is at 37.0° clockwise from the vertical, and the axis of the third polarizing filter is at 75.0° clockwise from the vertical. What is the intensity of the light after it has passed through all three polarizing filters?
B) 1.39 W/m2
C) 2.17 W/m2
D) 2.76 W/m2
E) 3.83 W/m2
F) none of the above answers
Unpolarized light of initial intensity 26.0 W/m 2 is passed through three polarizing filters. Viewed in the direction the light is traveling, the axis of the first polarizing filter is vertical, the axis of the second polarizing filter is at 35.0° clockwise from the vertical, and the axis of the third polarizing filter is at 75.0° clockwise from the vertical. What is the intensity of the light after it has passed through all three polarizing filters?
Unpolarized light with intensity I0 is passed two polarizing filters. The axis of the first filter is horizontal and the axis of the second filter makes an angle of 36.9° with respect to the horizontal. The intensity of the light after it has passed through the second filter is
A) 0.32 I0
B) 0.36 I0
C) 0.40 I0
D) 0.50 I0
E) 0.80 I0
F) None of the above answers
Light that is initially unpolarized and with initial intensity I0 passes through two polarizing filters. The angle between the polarizing axis of the first and second filter is 45°. What is the intensity of the light after it has passes through the second filter?
(b) 0.71 I0
(c) 0.50 I0
(d) 0.25 I0
(f) none of the above answers
Light with intensity I0 = 300 W/m2 passes through two ideal polarizing filters, which are oriented as shown in the figure. The angle of the first polarizer is fixed, while the angle Φ of the second filter is adjustable.
a) If the initial light is unpolarized, what value of Φ will make the intensity at point P equal to 100 W/m2?
Φ = _______________
b) Assuming the same conditions as in part (a), what are the maximum electric and magnetic field values at point P? (ε0 = 8.854 x 10-12 C2 / (N•m2))
Emax = _______________
Bmax = _______________
c) Now assume the inital light is linearly polarized in the same direction as the polarizing axis of the first polarizer. If Φ is fixed at the angle found in part (a), what will the intensity at point P be?
I = _______________
Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 19% of the light passes through, what is the measure of the angle between the transmission axes of the filters?