Polarized sunglasses at 17 degrees to vertical, reflected light 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.