Fresnel Equations for the Power of Reflectance and Transmittance

In summary, the conversation discusses the relationship between the power of reflectance and transmittance for both S-polarized and P-polarized light. The equations for this relationship are provided and the question of how to confirm this relationship is discussed. The process involves using trig substitutions and finding alternate forms of trig expressions, ultimately resulting in the confirmation that both R+T=1 for both types of light.
  • #1
Athenian
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Homework Statement
Check for ##R + T = 1## for both S-polarized and P-polarized light.
Relevant Equations
S-polarized Light
$$R_s + T_s = 1 \Longrightarrow \frac{sin^2 (\theta_t - \theta_i)}{sin^2 (\theta_t + \theta_i)} + \frac{sin(2\theta_i) sin(2\theta_t)}{sin^2 (\theta_t + \theta_i)} = 1$$

P-polarized light
$$R_p + T_p = 1 \Longrightarrow \frac{tan^2 (\theta_t - \theta_i)}{tan^2 (\theta_t + \theta_i)} + \frac{sin(2\theta_i) sin(2 \theta_t)}{sin^2 (\theta_t + \theta_i) cos^2 (\theta_i - \theta_t)} = 1$$
I understand that the power of reflectance and the power of transmittance (regardless of it being S-polarized or P-polarized light), when added together, would equal to one. In other words, ##R+T=1##. However, how do I go about checking for this fact?

Using the equations from the "Relevant Equations" above, I find that ##R_s + T_s = 1## or ##\frac{sin^2 (\theta_t - \theta_i)}{sin^2 (\theta_t + \theta_i)} + \frac{sin(2\theta_i) sin(2\theta_t)}{sin^2 (\theta_t + \theta_i)} = 1##. Now, how do check if this is indeed true rather than simply believing it to be true without checking?

Similarly, I know that ##R_p + T_p = 1## (for P-polarized light). And, I also know that through expanding the equation that ##\frac{tan^2 (\theta_t - \theta_i)}{tan^2 (\theta_t + \theta_i)} + \frac{sin(2\theta_i) sin(2 \theta_t)}{sin^2 (\theta_t + \theta_i) cos^2 (\theta_i - \theta_t)} = 1##. Likewise, though, how do I check and confirm this as true?

Any assistance and hints to help me understand how to go about solving this question would be greatly appreciated. Thank you for reading this post!
 
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  • #2
Never mind. I was able to finally solve this. To make a long calculation process simplified with a couple of words, the process involved A LOT of trig substitutions or finding alternate forms of a trig expression. In the end, one would find that both the numerator and the denominator contain the same expression, thus equalling to 1.
 

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