TBR light polarization

[PanRoasted]

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I don't quite understand the example 10.1a question in the TBR physics book (c.2009), or how polarization of light works in general (but lets just start with that question). Also, if someone could explain to me the example 10.1b question as well, i would be forever grateful.

the part about polarization I don't really understand is what exactly it does, and what implications this has. I mean, just from reading the section on polarization I get that the vectors radiate in only one plane, but what of the electric/magnetic waves? So only one component of any given wave can move through a polarizing material? Also, how does this affect anything? light is just a very confusing topic for me in general, i'm hoping it won't be represented too much on the MCAT i'll end up taking.

 

[chrisharimlee]

For 10.1a
you just need to understand v=wavelength x f. The question tells you that period is fixed, so that means f is fixed as well because T=1/f. So the question is if you decrease v what will happen? This means that your wavelength will have to decrease. However because f is fixed, the energy is also fixed, due to equation E=hf where h is a constant and E is the energy of a photon.

 

[chrisharimlee]

is the question the same i have the 2012 version

 

[MedGrl@2022]

I don't quite understand the example 10.1a question in the TBR physics book (c.2009), or how polarization of light works in general (but lets just start with that question). Also, if someone could explain to me the example 10.1b question as well, i would be forever grateful.

the part about polarization I don't really understand is what exactly it does, and what implications this has. I mean, just from reading the section on polarization I get that the vectors radiate in only one plane, but what of the electric/magnetic waves? So only one component of any given wave can move through a polarizing material? Also, how does this affect anything? light is just a very confusing topic for me in general, i'm hoping it won't be represented too much on the MCAT i'll end up taking.

Hey I just figured out 10.1b. This website really helps. See the polarization of reflected light section. http://www.physicsclassroom.com/class/light/u12l1e.cfm#trans

I am still having issues with 10.1a. I see how they got to 60degrees but then how did they get to 30degrees. Have you or anyone else figured it out?

Here is the problem and Berkeley Review's solution:

Example 10.1a

Vertically polarized light is sent through an empty sample cell and then into a
horizontally oriented polarizer. No light gets through the polarizer. The
experiment is repeated, this time with a dilute solution of D-Glucose in the cell,
and one-quarter of the incident light intensity gets through the polarizer. What
must be TRUE of the solution and the rotation of the polarized light? (Assume
that the solution transmits all the incident light to the polarizer.)

A. The solution is optically active and rotates the polarization by 30degrees.
B. The solution is optically active and rotates the polarization by 45degrees.
C. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 60degrees.
D. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 90degrees.

Solution

We need to think about the optical properties of the solution's molecules and the
polarization rotation. First, optically active materials are materials that can alter
the polarization of light passing through them. This activity usually arises from
an asymmetry in the structure and/ or orientation of the molecules that make up
the material. Here, the solution is optically active--the activity arises from the
helical, and therefore, asymmetric structure of the D-Glucose molecules. This
might be illuminating, but it doesn't help to isolate the correct choice. We must
also think about the polarization rotation.

The light that does get through the polarizer is one-quarter as intense as the light
incident upon it. To calculate the amount of linearly polarized light getting
through a polarizer, use Malus' law:

I=(Io)cos^2(theta)

where I is the intensity of light getting through the polarizer, when an initial
intensity Io is incident upon the polarizer. The angle (theta) is the angle between the
light's polarization direction and the polarization axis of the polarizer. Here,

I/(Io)=1/4=cos^2(theta)

Solving for (theta) gives 60 degrees. Since the polarizer's polarization axis is horizontally
oriented, the exiting light must be at an angle 30 degrees to the vertical. In other words,
the solution rotates the polarization of the initially vertical light by 30 degrees.

The correct choice is A.

I don't understand that which is colored in red. Maybe I need a picture... :-/

 

[sdm33]

Know couple of things.
1. Only transverse wave get polarized , not longitudinal waves
2. It is the electric field that oscillates in one direction when polarizing light.
3. Maximum amount of light emerges, when pair of polarizing filters at 0 or 180 degrees angle with each other. Eg ; 180 – theta =
4. Minimum light emerge (zero intensity) with the pair of polarizing filters at 90 degrees angle with each other.

 

[MedGrl@2022]

Know couple of things.
1. Only transverse wave get polarized , not longitudinal waves
2. It is the electric field that oscillates in one direction when polarizing light.
3. Maximum amount of light emerges, when pair of polarizing filters at 0 or 180 degrees angle with each other. Eg ; 180 – theta =
4. Minimum light emerge (zero intensity) with the pair of polarizing filters at 90 degrees angle with each other.

How did 60 degrees become 30 degrees? I still do not quite get that bit.

Thank you for your help! 🙂

 

[circulus vitios]

How did 60 degrees become 30 degrees? I still do not quite get that bit.

Thank you for your help! 🙂

It took me a while to figure it out but it's a matter of reference.

The light's path is as follows:

Source (vertically polarized) --> solution --> horizontal polarizer

I take it you understand the calculation that gave you a rotation value of 60 degrees. This means that the light passes through the horizontal polarizer and is rotated 60 degrees with respect to the horizontal polarizer. The question wants to know the rotation of the light after passing through the solution. Remember that the source is vertically polarized so we find the angle with respect to the vertical polarizer's axis.

 

[ShySpliceosome]

Wow, I do not understand this at all ...where did the 60 degrees come from??

 

[FeinMS]

Wow, I do not understand this at all ...where did the 60 degrees come from??

It came from the fact that the light that passed the polarizer had intensity of 1/4 of the original light. There's Malus's Law which relates intensiy of light that went through a polarizer and the original light intensity. I=Io*cos(theta)^2. Where Io is the intensity of the original light. I is the intensity of light that went through the polarizer. theta is the angle made between the original light and polarizer. I=Io/4=Io*cos(theta)^2 gives us 1/4=cos(theta)^2. Solving for theta gives 60 degrees. 60 degrees is the angle between light rotated by the solution and horizontal polarizer. So essentially the solution rotates light by 90-60 as mentioned above.

 

[ShySpliceosome]

It came from the fact that the light that passed the polarizer had intensity of 1/4 of the original light. There's Malus's Law which relates intensiy of light that went through a polarizer and the original light intensity. I=Io*cos(theta)^2. Where Io is the intensity of the original light. I is the intensity of light that went through the polarizer. theta is the angle made between the original light and polarizer. I=Io/4=Io*cos(theta)^2 gives us 1/4=cos(theta)^2. Solving for theta gives 60 degrees. 60 degrees is the angle between light rotated by the solution and horizontal polarizer. So essentially the solution rotates light by 90-60 as mentioned above.

Okay, I understand it now...thanks for your help!!

 

[gettheleadout]

Know couple of things.
1. Only transverse wave get polarized , not longitudinal waves
2. It is the electric field that oscillates in one direction when polarizing light.
3. Maximum amount of light emerges, when pair of polarizing filters at 0 or 180 degrees angle with each other. Eg ; 180 – theta =
4. Minimum light emerge (zero intensity) with the pair of polarizing filters at 90 degrees angle with each other.

This is an old post in this thread, but this is misleading. The magnetic field component of polarized light also oscillates in a single plane, which is perpendicular to the plane of the electric field, but we simply define the axis of polarization based on the orientation of the electric field oscillation because we had to pick one.