How to Calculate Reflection Coefficient for Light at Air to Silver Interface?

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To calculate the reflection coefficient for light at the air-silver interface, one must consider the optical frequency and the conductivity of silver, which is approximately 6 x 10^7 mhom^{-1}. The reflection is nearly 100% due to the high conductivity of silver, and the reflected intensity can be expressed as I_r = I_0 - I_absorbed, where I_absorbed is calculated using I_{absorbed} = √(ω/8πσ)I_0 in Gaussian units. The impedance of silver (η_s) can be derived from the wavelength and conductivity, allowing for the calculation of the reflection coefficient using the formula r = (1 - β)/(1 + β), where β is the ratio of the impedance of free space (η_0) to that of silver. Overall, the approach discussed is valid for determining the reflection coefficient at this interface.
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How does one calculate the reflection coefficient for light at an air to silver interface at optical frequencies (\omega = {4 *10^{15} s^{-1}})
given

\mu_{I} = \mu_{T} = \mu_{0} , \sigma = {6 * 10^7 mhom^{-1}}
 
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The reflection is close to 100% for silver.
the reflected intensity is I_r=I_0-I_absorbed.
I_{absorbed}=\sqrt{\omega/8\pi\sigma}I_0 in Gaussian units.
 
Would this work?

Knowing the frequency means we can find the wavelength clearly, which has units metres, clearly. When combined with \sigma = {6 * 10^7 mhom^{-1}} to get the impedance of the silver. Then taking the ratio with the impedance of free space and assuming normal incidence

r =\frac{1 - \beta}{1 + \beta}

where \beta = \frac{\eta_0}{\eta_s}

Would this work?
 
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How would you calculate \eta?
 
eta_s is the impeance of silver. As indicated previously a value with ohm units can be achieved by combining wavelength with the value of conductivity. What I am asking is, is this a valid statement?
 
No. Units are not a calculation.
 
i want to know about the motion of the stelites totely
 
basic defination of electromagnatic physics
 

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