Index of Refraction, Snell's Law, and Brewster's Angle

[alibert0914]

Homework Statement

This was for a lab experiment, and I'm still not sure how this all fits together. We were supposed to use snell's law to find the index of refraction for the lab bench. We measured from the top of the bench to our eye level, and to the center of the bright spot seen through a polaraizer. Height was 0.741 m, and distance was 0.605 m. We used the ref. index of air for n1.

Homework Equations

tan (theta): height/distance

n1sin(theta1) = n2sin(theta2)

thetaB = arctan n2/n1

The Attempt at a Solution

We calculated an angle to be 50.9 degrees. We assumed this to be thetaB, and plugged it in. However, since we were using n1=1 for air, it just seems like a lot of back and forth, and we basically get arctan (thetaB) = n2= 1.23. Most everyone else in the lab seems to have gotten n2 to be less than one. If someone could give us a nudge in the right direction that would be very helpful. Thanks a lot!

 

[gneill]

An index of refraction less than 1.0 is rather unlikely I would think.

 

[cepheid]

Welcome to PF alibert0914!

I don't think you really need the equation for Snell's law, since refraction is not happening here (EDIT: for the polarization of interest, anyway). If memory serves, the Brewster angle is the angle at which light (with a specific polarization) that is incident upon an interface between two media will be totally reflected (i.e. none of it will be transmitted from the first medium to the second). Since the angle of incidence equals the angle of reflection, the angle you measured was equal to the angle of incidence. So you have

tan(θ_B) = n₂/n₁ = n₂

= height/distance = 1.22

As for your classmates -- a refractive index less than 1? I think not. This would mean that the speed of light in the lab bench material would be faster than c, the speed of light in a vacuum.

 

[alibert0914]

Thanks so much everyone! My partner and I just wanted to double check everything before we turn our report in.