Polarized Light Reflection: Solving for Intensity

[mli273]

1. A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical.

If the wearer leans at an angle of 17.0 degrees to the vertical, what fraction of the reflected light intensity will pass through the sunglasses?


2. I =I(initial)cos^2(theta)



3. I tried I/I(initial)=cos^2(17) which yielded .91 but that was not correct. What factor am I missing here?

 

[DaveC426913]

1. A person riding in a boat observes that the sunlight reflected by the water is polarized parallel to the surface of the water. The person is wearing polarized sunglasses with the polarization axis vertical.

If the wearer leans at an angle of 17.0 degrees to the vertical, what fraction of the reflected light intensity will pass through the sunglasses?


2. I =I(initial)cos^2(theta)



3. I tried I/I(initial)=cos^2(17) which yielded .91 but that was not correct. What factor am I missing here?

If the person is vertical, zero gets through, right?
If they lean just a leeeetle bit to the side, how much should get through?

9/10^ths?

Intuitively, does that make sense? Intuitively, what does make sense?

 

[mli273]

that's true, I didn't think of it like that. So would it be 1-.915 which is .085. Thank you!

 

[DaveC426913]

Never trust a calculator. Use and trust your common sense. The tool just gets you the decimals.

IMO, that's the one big lesson students need to learn.

 

[rwisz]

I would like to first clarify that polarized light is a type of light where the electromagnetic waves vibrate in a single direction. This is in contrast to unpolarized light, where the waves vibrate in all directions. Polarized sunglasses are designed to block out the horizontally polarized light, which is typically the glare from surfaces such as water or road.

Now, to solve for the intensity of the reflected light passing through the polarized sunglasses, we need to use the equation provided in the content: I = I(initial)cos^2(theta). In this equation, I represents the intensity of the reflected light after passing through the sunglasses, I(initial) represents the initial intensity of the reflected light, and theta represents the angle between the polarization axis of the sunglasses and the direction of the reflected light.

In this scenario, the wearer is leaning at an angle of 17.0 degrees to the vertical, which means the angle between the polarization axis of the sunglasses and the direction of the reflected light is also 17.0 degrees. Therefore, we can rewrite the equation as follows: I = I(initial)cos^2(17.0).

To solve for the fraction of the reflected light intensity that will pass through the sunglasses, we need to divide the intensity after passing through the sunglasses (I) by the initial intensity (I(initial)). This can be written as I/I(initial) = cos^2(17.0). Using a calculator, we can find that cos^2(17.0) is approximately 0.91, which means that 91% of the reflected light intensity will pass through the sunglasses.

In conclusion, the factor that was missing in your calculation was the angle between the polarization axis of the sunglasses and the direction of the reflected light. By including this angle in the equation, we can accurately solve for the fraction of the reflected light intensity that will pass through the polarized sunglasses.