- #1
binbagsss
- 1,326
- 12
The question asks to use Fresnel's equations at normal incidence to find the ratio of the refractive indexes of the two materials. We are told equal energy is transmitted and reflected and that both media are non-magnetic and non-conducting dielectrics.
So I know that the ratio of Er/Ei and Et/Ei are the same for this case, regardless of whether the electromagnetic wave is parallel or perpendicular to the planar interface .
In this case, we are also told both media are non-magnetic. So the ratios of Er/Ei and Et/Ei simplify further as ui=ut=u0
(where Ei is the incident electric field , Er reflected, ui the permeability of the media with the incident beam etc.)
Now I am then asked if one of the media has a relative permittivity of 16 ,what is the refractive index of the other media?
This is probably a stupid question, but if non-magnetic means to take the permeability to be the permeability of free space, then doesn't non-conducting mean to take the permittivity to be the permittivy of free space?
And n = ([itex]\epsilon_{r}[/itex][itex]\mu_{r}[/itex])[itex]^{\frac{1}{2}}[/itex]. So they would both be equal and equal 1 which is obviously wrong.
( I am able to follow the working to get the correct expression for the ratio of the refractive indexes, I'm just missing some key concepts clearly...)
Many thanks for any assistance !
So I know that the ratio of Er/Ei and Et/Ei are the same for this case, regardless of whether the electromagnetic wave is parallel or perpendicular to the planar interface .
In this case, we are also told both media are non-magnetic. So the ratios of Er/Ei and Et/Ei simplify further as ui=ut=u0
(where Ei is the incident electric field , Er reflected, ui the permeability of the media with the incident beam etc.)
Now I am then asked if one of the media has a relative permittivity of 16 ,what is the refractive index of the other media?
This is probably a stupid question, but if non-magnetic means to take the permeability to be the permeability of free space, then doesn't non-conducting mean to take the permittivity to be the permittivy of free space?
And n = ([itex]\epsilon_{r}[/itex][itex]\mu_{r}[/itex])[itex]^{\frac{1}{2}}[/itex]. So they would both be equal and equal 1 which is obviously wrong.
( I am able to follow the working to get the correct expression for the ratio of the refractive indexes, I'm just missing some key concepts clearly...)
Many thanks for any assistance !