Correct statement about light transmitted and reflected by windscreen

[songoku]

Homework Statement

Please see below

Relevant Equations

not sure

I don't really understand how to relate the question to the picture. Picture Y is taken using polarising filter but how can we know whether it is light transmitted or reflected because I think in both pictures there is light transmitted through the windscreen and there is also light reflected from the windscreen.

Thanks

 

[kuruman]

I don't really understand how to relate the question to the picture. Picture Y is taken using polarising filter but how can we know whether it is light transmitted or reflected because I think in both pictures there is light transmitted through the windscreen and there is also light reflected from the windscreen.

Just compare the two images. Yes, there is light transmitted through and light reflected by the windshield. Can you sort out which is which? Study the two pictures carefully and look for clues.

 

[songoku]

Just compare the two images. Yes, there is light transmitted through and light reflected by the windshield. Can you sort out which is which? Study the two pictures carefully and look for clues.

The only clue I can think of is from picture X we can't see the inside of the car but we can from picture Y. The reason will be probably because using polarising filter the glare can be removed so the image will be clearer but I still can't sort out which one is dealing with light transmitted and which one is light reflected.

Thanks

 

[renormalize]

The only clue I can think of is from picture X we can't see the inside of the car but we can from picture Y.

So in picture Y the light from inside the car gets through the windshield and the polarizing filter, whereas the light from the sky that's reflected by the windshield is blocked by the filter and hence the glare disappears. What does that tell you?

 

[kuruman]

The only clue I can think of is from picture X we can't see the inside of the car but we can from picture Y. The reason will be probably because using polarising filter the glare can be removed so the image will be clearer but I still can't sort out which one is dealing with light transmitted and which one is light reflected.

Don't say glare, say reflection off the windshield. The tree branch seen across the windshield from right to left is not glare and is seen in both photographs. That, too, needs explanation.

 

[Charles Link]

I do think the choices in the table aren't as clear as they could be, and they might do better to call the "unpolarized"case perpendicularly polarized.The concept they are trying to teach is basically all about the Brewster angle, and how the graph of reflectivity vs. angle is different for the polarization parallel to the plane vs. perpendicular to the plane. The term "plane"polarized IMO is rather ambiguous.

Suggestion for the OP is to sketch the graph of reflectivity vs. angle for the parallel polarization case, and for the perpendicularly polarized case for reflection off a material such as glass with index $n=1.5$. You should then see that the reflected light anywhere near the Brewster angle will be largely polarized perpendicular to the plane of incidence. I looked for a graph of these two cases on-line, but I couldn't find a good one. The parallel case has the reflectivity go to zero at the Brewster angle and then increases past the Brewster angle, while the perpendicular case increases steadily with angle of incidence with no Brewster angle. For both directions of polarization the reflectivity goes to $1.0$ at 90 degree angle of incidence. (When it is polarized perpendicular to the plane of incidence and reflection it is certainly polarized, but the term "plane" polarized to me is unclear and IMO is somewhat confusing. I don't know if you would call the perpendicular case "plane" polarized.)

 

[Charles Link]

One other thing they omit=I hope they know the subject they are teaching well enough, is that the polarizing filter needs to have its direction of polarization properly oriented. It isn't just a photo taken with a polarizing filter. It is a photo taken with a polarizing filter properly oriented.

With polaroid sunglasses, the polarization direction of the sunglasses you wear is with polarization (passing) up and down, i.e. vertically. Most glare will be polarized perpendicular to the up and down direction, i.e. horizontally, and thereby the sunglasses will block the glare.

Rotate your polaroid sunglasses 90 degrees, and they will not block the glare.

 

[songoku]

I am sorry for late reply

So in picture Y the light from inside the car gets through the windshield and the polarizing filter, whereas the light from the sky that's reflected by the windshield is blocked by the filter and hence the glare disappears. What does that tell you?

In picture X, instead of the inside of the car, we can see the sky but in picture Y most of the sky can not be seen and we can see the inside of the car so this means the light from the sky that is reflected by the windscreen is being blocked by the filter.

For the filter to be able to block the light, it means the light reflected from the windshield must be polarised. But I am not sure about the light transmitted through windscreen. For this light to be able to pass through the polarising filter, I think it can be either unpolarised or polarised (as long as the polarization is not perpendicular to the filter). So for safe measure, we just take the light transmitted through windscreen is unpolarised?

Don't say glare, say reflection off the windshield. The tree branch seen across the windshield from right to left is not glare and is seen in both photographs. That, too, needs explanation.

I am also not sure about the tree branch. Does it show that light from the sky has polarization perpendicular to the filter and the light reflected from the tree branch and then from the windscreen has polarization parallel to the filter?

I do think the choices in the table aren't as clear as they could be, and they might do better to call the "unpolarized"case perpendicularly polarized.The concept they are trying to teach is basically all about the Brewster angle, and how the graph of reflectivity vs. angle is different for the polarization parallel to the plane vs. perpendicular to the plane. The term "plane"polarized IMO is rather ambiguous.

Suggestion for the OP is to sketch the graph of reflectivity vs. angle for the parallel polarization case, and for the perpendicularly polarized case for reflection off a material such as glass with index $n=1.5$. You should then see that the reflected light anywhere near the Brewster angle will be largely polarized perpendicular to the plane of incidence. I looked for a graph of these two cases on-line, but I couldn't find a good one. The parallel case has the reflectivity go to zero at the Brewster angle and then increases past the Brewster angle, while the perpendicular case increases steadily with angle of incidence with no Brewster angle. For both directions of polarization the reflectivity goes to $1.0$ at 90 degree angle of incidence. (When it is polarized perpendicular to the plane of incidence and reflection it is certainly polarized, but the term "plane" polarized to me is unclear and IMO is somewhat confusing. I don't know if you would call the perpendicular case "plane" polarized.)

I am really sorry I don't understand how to sketch the graph you are referring to

 

[Charles Link]

I am really sorry I don't understand how to sketch the graph you are referring to

I wish someone could find this graph in a "link" somewhere or sketch it, (reflectivity vs. angle for the parallel and perpendicular polarization), because it really shows how reflected light is polarized, and even how the transmitted light through a dielectric tilted at an angle will then also be partially polarized, but with the polarization at 90 degrees to that of the reflected light's polarization.

Edit: I found it. See https://physics.stackexchange.com/q...nsely-when-it-hits-a-surface-at-a-large-angle

The choices available in the problem statement didn't seem very clear to me, because calling it polarized or unpolarized light isn't really descriptive enough. The direction of the polarization, e.g. vertical or horizontal needs to be specified, just as the polarizing filter that was placed in front of the camera lens. The transmitted light through the windshield will be partially polarized in the vertical direction. The light reflected off the windshield will be partially polarized horizontally. (The extent of the polarization will increase as the angle of incidence approaches the Brewster angle). The filter on the camera lens is polarized in the vertical direction, which they really should have stated in the problem statement. I really don't know what your instructor will say is the correct answer to the problem.

Unpolarized light will get polarized as it passes through a polarizing filter and will then be polarized in the direction that the filter is oriented. Its intensity will be reduced by 50%.

Light that is horizontally polarized will not pass through a polarizing filter that is oriented in the vertical direction. Light that is polarized in the vertical direction will pass through a polarizing filter that is oriented in the vertical direction. See https://www.geeksforgeeks.org/malus-law/

 

[songoku]

I wish someone could find this graph in a "link" somewhere or sketch it, (reflectivity vs. angle for the parallel and perpendicular polarization), because it really shows how reflected light is polarized, and even how the transmitted light through a dielectric tilted at an angle will then also be partially polarized, but with the polarization at 90 degrees to that of the reflected light's polarization.

Edit: I found it. See https://physics.stackexchange.com/q...nsely-when-it-hits-a-surface-at-a-large-angle

In the link, there is this graph:

What is Rs and Rp?

The transmitted light through the windshield will be partially polarized in the vertical direction. The light reflected off the windshield will be partially polarized horizontally. (The extent of the polarization will increase as the angle of incidence approaches the Brewster angle). The filter on the camera lens is polarized in the vertical direction, which they really should have stated in the problem statement.

How can we know the orientation of the polarization? Is it always like that; reflected will be horizontal, transmitted will be vertical and polarising filter is vertical?

I really don't know what your instructor will say is the correct answer to the problem.

I asked and he said the answer is (C).


Thanks

 

[Charles Link]

There is a plane defined by the incident ray, the reflected ray, and the normal to the surface. I think this is usually referred to as the plane of reflection. Light can be polarized parallel to this plane, or it can be polarized perpendicular to this plane. The two components behave differently (in intensity) upon reflection. The parallel component ,$R_p$, has an angle where the reflectivity drops to zero, known as the Brewster angle. Meanwhile the reflectivity of the perpendicular component $R_s$ increases steadily with angle of incidence.

The windshield of the vehicle is tilted somewhere near the Brewster angle. Light reflecting off of it will be polarized in the perpendicular direction for the most part, which is horizontal in this case. The transmitted light will be a combination of both types of polarization, but somewhat more of the parallel (vertical) variety. They really should have told the student that the polarizing filter in front of the camera is oriented in the vertical direction, i.e. to pass the parallel polarized light and block the perpendicular.

I don't know that (C) is the best answer, but I thought they should have specified for the polarized choices the direction of polarization. Just to call it polarized is insufficient, and unless you know the direction, you don't know if the filter will pass or block it. (Meanwhile the polarizing filter can be turned at some arbitrary angle, but we are just considering vertical and horizontal cases here for now).

 

[songoku]

Thank you very much for the help and explanation kuruman, renormalize, Charles Link