Intensity of Light by Polarizers

In summary, the l-leucine solution has a .550o rotation while the d-glutamic acid solution has a .620o rotation.
  • #1
deenuh20
50
0

Homework Statement




Some biologically active molecules rotate the direction of polarization of linearly polarized light, either clockwise or anticlockwise. For example, 5 gm of l-leucine dissolved in 100 mL of water causes a rotation by 0.550o while the same concentration of d-glutamic acid causes a rotation of 0.620o. If these solutions are placed between crossed polarizers illuminated by a beam of unpolarized light of intensity I=13 W/m2, what is the intensity of light transmitted by:

(a) the l-leucine solution? ___W/m^2
(b) the d-glutamic acid solution? ___W/m^2

Homework Equations


I=Io*cos^2(theta)

I=.5*Io


The Attempt at a Solution



To do this, for each solution, I plugged in the numbers in the equation:

for l-leucine: I=(13 W/m^2)(cos^2(0.62))
for d-glutamate: I=(13 W/m^2)(cos^2(0.55))

That didn't work. So I then tried finding I by doing:
I=(.5)(13)=6.75 then used this is the above equation and tried solving for Io then. That didn't work. A hint or suggestion would be greatly appreciated. Thank you!
 
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  • #2
Hi deenuh,
.55 degrees and .62 degrees are extremely small angles and the cosine of these angles squared will be much closer to one than the numbers which you are using.

I hope that helps.
 
  • #3
Take a look at Malus Law again:
Try to explain what I=Io*cos^2(theta) means.
It helps if you first consider polarized light that is incident on a polarizer.

What is theta?
What is Io?
 
Last edited:
  • #4
Isn't Io equal to the beam of unpolarized light (in this case, 13 W/m^2)? Wouldn't theta be 45 degrees because the polarizers are crossed?
 
  • #5
Crossed filters are at a 90 degree angle. If the samples didnt rotate the polarisation of the light none would get through.

You are going through two filters. First unpolarised light goes through a filter. Youve given the formula that tells us the intensity drop for this. Then this polarised light goes through a second filter which is 90 degrees to the first. Youve given the right formula for this as well you just need to think again about what is the right value for theta and Io for each sample. Theta is the angle between the plane of polarisation of the light and the plane of polarisation of the filter.
 
  • #6
Thanks for your suggestions, I was finally able to calculate the answer!
 

1. What is the concept of intensity of light by polarizers?

The intensity of light by polarizers refers to the amount of light that passes through a polarizing filter or polarizer. This filter is able to block certain polarizations of light, resulting in a reduction of the intensity of the light that passes through it.

2. How does a polarizer affect the intensity of light?

A polarizer works by only allowing light waves with a certain polarization to pass through it. This means that the intensity of light passing through the polarizer is reduced, as only a portion of the light is able to pass through.

3. What factors affect the intensity of light by polarizers?

The intensity of light by polarizers can be affected by the angle of the polarizer, the intensity of the incident light, and the orientation of the polarizer relative to the light source. Additionally, the type and thickness of the polarizing material can also impact the intensity of light.

4. How is the intensity of light measured by polarizers?

The intensity of light by polarizers can be measured using a light intensity meter, also known as a photometer. This instrument measures the amount of light passing through a polarizer and can provide a numerical value for the intensity of light.

5. Can the intensity of light be completely blocked by a polarizer?

No, the intensity of light cannot be completely blocked by a polarizer. Even in the most efficient polarizers, a small amount of light is still able to pass through due to imperfections in the polarizing material. However, the intensity of light can be significantly reduced by using multiple polarizers or by adjusting the angle of the polarizer.

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