Sunlight contains equal power densities in both the parallel and perpendicular polarizations. Assume that sunlight is incident at a 40 angle on a puddle of water. Because the water is relatively pure fresh water, it is nonconductive and has a dielect constant (relative permittivity) of epsilon_r = 1.78. The water is nonmagnetic What is the percentage of power that is reflected?

Answer / Explanation

To start with in answering this question, let us first recall the equation used in calculating the percentage of reflected power:

That is:

Reflected Power = %Pref = 100 ║Γ║²let this be equation (1)

Taking note from the expression above that:

Γ =  Reflection Coefficient.

The reflection coefficient is the variable that shows the level of electromagnetic wave that is reflected by an impedance break in the transmission medium.

Now, recalling the expression in finding the reflection coefficient,

we have: Γ = η₂ - η₁ / η₂ +η₁ let this be equation (2)

Where:

η₁ and η₂ = intrinsic Impedance

Now, if we recall the relationship between intrinsic impedance and relative permitivities,

We have: η₂ / η₁ = √ εr₁ / εr₂

Now, if we refer back to the narrative of the question, we will substitute 1.78 for  εr₂ and substitute 1 for εr₁

So, moving forward, we have:

η₂ / η₁ = √ 1 / 1.78

η₂ / η₁  = 0.7495

η₂ = 0.7495 η₁ let this be equation (3)

Now, moving forward, we will substitute equation (3) into equation (2)

Therefore, we have:

Γ = 0.7495 η₁ - η₁ / 0.7495 η₁ + η₁

Γ = - 0.2505η₁ / 1.7495η₁

Γ = - 0.1431

Now, we go ahead to substitute the value of the reflection coefficient into equation (1)

Thus, we have:

Recalling equation (1):

%Pref = 100 ║Γ║²

%Pref = 100 x ( 0.1431) ²

= 2.0477%

Approximating the value above, we will have:

%Pref = 2.05%

Therefore, we  can conclude that the percentage of power that is reflected is 2.05%

me and ten points now

Explanation: