Question

A beam of natural light is incident on an air-glass interface \((n_{ti}=1.5)\) at \(40^{\circ}\). Compute the degree of polarization of the reflected light.

Answer 1

I calculated \(R_\perp=r_\perp^2\) and \(R_\parallel=r^2_\parallel\) where \[R_\parallel=\frac{\sin^2 (\theta_i-\theta_t)}{\sin^2(\theta_i+\theta_t)}\]\[R_\perp=\frac{\tan^2 (\theta_i-\theta_t)}{\tan^2(\theta_i+\theta_t)}\] and using snell's law for \(\theta_t\)

Answer 2

so \[R=\frac{R_\perp+R_\parallel}{2}\]now my formula for the degree of polarization is \[V=\frac{I_p}{I_p+I_n}\]

Answer 3

so my textbook says this is equvalent to \[\frac{R_\perp+R_\parallel}{R_\perp+R_\parallel+R}\]but my prof crossed that out and used \[\frac{I_{\text{max}}-I_{\text{min}}}{I_{\text{max}}+I_{\text{min}}}\]

Answer 4

so my question is how do I find \(I_\text{max}\) and \(I_\text{min}\)?