Calculating Reflective coefficient for glass slabby jlefevre76 Tags: incident angle, index of refraction, optics, reflection, reflectivity 

#1
Mar1813, 01:16 PM

P: 23

Okay, so now I've lost all confidence in myself. I've had a lot of issues calculating the reflective coefficient for a glass slab and a thin film solar cell. I know this method isn't perfect, and not as good as experimental data (which, if anyone here can supply, I'd be happy to take it). However, I need to get my raytracing analysis done, and hopefully get published again if possible. These calculations are critical to me getting a somewhat valid answer for my raytracing calculation. (FYI this is research, not homework)
So, I'm referring to "Thermal Radiation Heat Transfer" by Howell, Siegel, and Menguc. Chapter 17 details the equations. I'm assuming an unpolarized source (the sun), so 50% is polarized one way, and 50% the other (correct me if wrong, anybody). So, this is how I THINK it's done: Calculate the refracted angle (θ_{glass}): n_{air}sin(θ_{incident}) = n_{glass}sin(θ_{glass}) θ_{glass} = arcsin(n_{air}sin(θ_{incident})/n_{glass}) Then, the surface reflectivity is calculated: ρ = (1/2)(tan^{2}(θ_{incident}θ_{glass})/tan^{2}(θ_{incident}+θ_{glass})+sin^{2}(θ_{incident}θ_{glass})/sin^{2}(θ_{incident}+θ_{glass})) Then, just to keep it simple this time, assume the transmissivity = 1 (I can also calculate this with an attenuation coefficient, but no need to include that part here): τ = 1 So, the reflective coefficient for a glass slab would be: R = ρ(1+((1ρ)^{2}τ^{2})/(1(ρτ)^{2})) R = ρ(1+(1ρ)^{2}/(1ρ^{2})) A similar equation comes up for the photovoltaic cells, and I think if I can get this one right, I can get the other one right. Part of my confusion for how the equation for the reflective coefficient is derived stems from the fact that they assume the reflectivity (ρ) is the same on both sides of the slab. Would this be the case? Because, in my mind, as it enters the glass, it's going from a low index of refraction to a high one. Then, as it hits the other air/glass interface, it's going from a high index to a low one. Is ρ really the same for each side as the ray passes through the glass? I've attached a spreadsheet that should show the results. Thanks in advance. Let me know of errors, etc. I'll try to correct them. 



#2
Mar2013, 03:45 PM

P: 23

Well, gonna go ahead and bump my own post. If I have done something wrong so I'm not eliciting a response, please let me know (I'm new, after all). Also let me know if I didn't explain something very well.




#3
Mar2513, 05:17 PM

P: 23

So, no answer from the forums, I decided to take matters into my own hands.
My supplies: A homemade UV LED flashlight, masking tape, a multimeter, alligator clips, a sheet of plexiglas, a CIS solar cell, and a tape measure. Well, probably the first datapoint or two would have been enough, but I did a few, with honestly, really good results for a garage experiment (well, or basement bathroom). There is some noise in the data, but it's good enough to give me confidence that I have correctly applied the theory. Take a look. Hope this helps somebody some time. 


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