Why Does Reflection Phase Shift Not Always Apply to Both Polarizations?

AI Thread Summary
The discussion centers on the phase shift of reflected light rays, specifically addressing why the common assertion that a phase shift of π occurs when n1 < n2 applies only to s-polarization. It is clarified that the p-polarization reflection coefficient only becomes negative under specific conditions, contrasting with s-polarization. The participants highlight that while many optics textbooks may imply a general rule, they often do not distinguish between polarizations. The conversation reveals a potential misunderstanding in the interpretation of phase shifts among learners, suggesting that the assumption of s-polarization may be prevalent in educational contexts. Overall, the nuances of reflection coefficients for different polarizations are crucial for a comprehensive understanding of light behavior at interfaces.
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Its always said that a reflected light ray acquires a phase shift equal to ## \pi ## if ## n_1 < n_2 ##. But considering the Fresnel coefficients, its revealed that its only for the s-polarization reflection coefficient that ## n_1 < n_2 ## causes the coefficient become negative. The p-polarization reflection coefficient becomes negative only when ## \sin^2 \theta_1 > \frac{1}{1+(\frac{n_1}{n_2})^2} ##. So why the first sentence doesn't distinguish different polarizations?
Thanks
 
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I guess the author implicitly assumes normal incidence.
 
I just found* that the p reflection coefficient becomes negative when ## n_2 < n_1 ##, exactly the opposite condition for s reflection coefficient!

*## n_2 \cos\theta_1<n_1 \cos\theta_2 \Rightarrow \frac{n_2}{n_1} \cos\theta_1 < \cos\theta_2 \Rightarrow \sin\theta_1\cos\theta_1 < \sin\theta_2 \cos\theta_2 \Rightarrow \sin{2\theta_1} < \sin{2 \theta_2} \Rightarrow \\ \theta_1 < \theta_2 \Rightarrow n_2 < n_1##
 
What does it have to do with the original problem?
 
It seems we have two conditions that result in ## r_p < 0## but only one for ## r_s<0 ##. So we have the following situations:
1) ## n_1 < n_2 ## and ## \sin^2{\theta_1}<\frac{1}{1+(\frac{n_1}{n_2})^2} ##: Only the s polarization shifts phase upon reflection.
2) ## n_1 < n_2 ## and ## \sin^2{\theta_1}>\frac{1}{1+(\frac{n_1}{n_2})^2} ##: Both polarizations shift phase upon reflection.
3) ## n_1 > n_2 ##: Only p polarization shifts phase upon reflection.
Well, at least now I have a clearer view. I'm beginning to think that the optics textbooks implicitly assume the light ray to have s polarization. Am I overestimating the number of textbooks that claim ## n_1 < n_2 ## means there is a phase shift upon reflection? Actually someone asked me this question and I remember in my own optics course that the professor kept repeating that there is phase shift when ## n_1 < n_2 ##. This is also abundant on the internet(this, this, this and this). But I don't remember whether textbooks claim as such or not. It seems to me that textbooks get it right but it became a misunderstanding among people.
 
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