Understanding Reflectance and Transmittance Calculations

[LCSphysicist]

Homework Statement

A beam of unpolarized light carries 2000W/m2 down onto an air–plastic interface. It is found that of the light reflected at the interface 300W/m2 is polarized with its E-field perpendicular to the plane of incidence and 200W/m2 parallel to the plane of incidence. Determine the net transmittance across the interface.

Relevant Equations

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This type of problem always confuse me. Indeed, i am always skeptical with the answer i get, because i know there is some subtle points in the calculations that maybe i am missing... Anyway:

Can we just say that $$T = 1 - R = 1 - \frac{(R_{\parallel}+R_{\bot})}{2} = 1 - \frac{(\frac{200}{2000}+\frac{300}{2000})}{2} = 1 - 5/40 = 35/40$$ ?

 

[kuruman]

I think you need to consider that the incidence may not be normal. The incident beam contains a 50-50 mix of light polarized in the plane of incidence (p-polarization) and perpendicular to it (s-polarization). The ratio of electric field magnitudes before and after reflection depends on the polarization and the angle of incidence because of the boundary conditions. Relevant are the Fresnel equations which should help you find the angle of incidence from the amount of reflected p and s polarization. Don't forget that the electric field magnitude is proportional to the square root of the energy flux.

Disclaimer: I have not solved this problem. I just described what I would try first.

 

[LCSphysicist]

I think you need to consider that the incidence may not be normal. The incident beam contains a 50-50 mix of light polarized in the plane of incidence (p-polarization) and perpendicular to it (s-polarization). The ratio of electric field magnitudes before and after reflection depends on the polarization and the angle of incidence because of the boundary conditions. Relevant are the Fresnel equations which should help you find the angle of incidence from the amount of reflected p and s polarization. Don't forget that the electric field magnitude is proportional to the square root of the energy flux.

Disclaimer: I have not solved this problem. I just described what I would try first.

Yes but, i didn't assumed that the incidence was normal (did i??)
That $R = \frac{R_{\parallel}+R_{\bot}}{2}$ is jut true to any incidence, as well is $T+R = 1$, being T defined analogous as R.
Or have i misinterpreted the theory?

(Note: i am not talking about $r,t$ (fraction of amplitudes), but $R,T$ (fraction of intensity))

 

[TSny]

Homework Statement:: A beam of unpolarized light carries 2000W/m2 down onto an air–plastic interface. It is found that of the light reflected at the interface 300W/m2 is polarized with its E-field perpendicular to the plane of incidence and 200W/m2 parallel to the plane of incidence. Determine the net transmittance across the interface.

Can we just say that $$T = 1 - R = 1 - \frac{(R_{\parallel}+R_{\bot})}{2} = 1 - \frac{(\frac{200}{2000}+\frac{300}{2000})}{2} = 1 - 5/40 = 35/40$$

Based on your result for $T_{\rm net}$, what is the net intensity transmitted in W/m²?

Does the net transmitted intensity plus the net reflected intensity equal the net incoming intensity (2000 W/m²)?

 

[TSny]

Can we just say that $$T = 1 - R = 1 - \frac{(R_{\parallel}+R_{\bot})}{2} = 1 - \frac{(\frac{200}{2000}+\frac{300}{2000})}{2} = 1 - 5/40 = 35/40$$

We have to be careful with how we calculate $R_{\parallel}$ and $R_{\bot}$.

In a handwaving way we can say that if we have 2000 W/m² of incident unpolarized light , then on the average we have 1000 W/m² incident with || polarization and 1000 W/m² incident with ⊥ polarization. So, $R_{\parallel} = \frac{200}{1000}$ and similarly for $R_{\bot}$.

 

[LCSphysicist]

We have to be careful with how we calculate $R_{\parallel}$ and $R_{\bot}$.

In a handwaving way we can say that if we have 2000 W/m² of incident unpolarized light , then on the average we have 1000 W/m² incident with || polarization and 1000 W/m² incident with ⊥ polarization. So, $R_{\parallel} = \frac{200}{1000}$ and similarly for $R_{\bot}$.

Nice. So we got $T = 1 - (200/1000 + 300/1000)/2 = 3/4$?

 

[TSny]

Nice. So we got $T = 1 - (200/1000 + 300/1000)/2 = 3/4$?

Yes, I think that's right.