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As explained in a previous reply there are values of wavelength for which n = 1.2999 for ice at -40 C, but the problem is the angles would most likely have to be different.

However, we can look at it this way. If we assume that n(2) = 1.2999, and that the angles are as given, we can calculate what the refractive index of the air must be to produce this result (i.e. n(1)).

We find that the refractive index of air must be 1.000329699. This implies that the wavelength of light must be greater than 1250 nm (since this is the highest i can calculate for you using the Mathematica program i linked to). The refractive index of air is less than 1.000333994 above 1250 nm. However, I may be able to find an approximate wavelength for you later as it by fitting data points and extrapolating to higher wavelengths. BUT the refractive index of air must be either 8222, 12109, 30826, or 46423.5 nm (we have excluded the 84.815 and 1095 values) since these are the wavelengths that produce a refractive index of ice at -40 C of n(2) = 1.2999. Thus, if the extrapolated wavelength is not at these values, then there is no wavelength that can be used to produce the required angles for air and ice at - 40C.

Jonny

This post was created using 100% recycled electrons

However, we can look at it this way. If we assume that n(2) = 1.2999, and that the angles are as given, we can calculate what the refractive index of the air must be to produce this result (i.e. n(1)).

We find that the refractive index of air must be 1.000329699. This implies that the wavelength of light must be greater than 1250 nm (since this is the highest i can calculate for you using the Mathematica program i linked to). The refractive index of air is less than 1.000333994 above 1250 nm. However, I may be able to find an approximate wavelength for you later as it by fitting data points and extrapolating to higher wavelengths. BUT the refractive index of air must be either 8222, 12109, 30826, or 46423.5 nm (we have excluded the 84.815 and 1095 values) since these are the wavelengths that produce a refractive index of ice at -40 C of n(2) = 1.2999. Thus, if the extrapolated wavelength is not at these values, then there is no wavelength that can be used to produce the required angles for air and ice at - 40C.

Jonny

This post was created using 100% recycled electrons

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