he refractive index and permeability of a medium are 1.51.5 and 5×10−7 H m−15×10-7 H m-1 respectively. The relative permittivity of the medium is nearly
Question
he refractive index and permeability of a medium are 1.51.5 and 5×10−7 H m−15×10-7 H m-1 respectively. The relative permittivity of the medium is nearly
Solution
The refractive index (n) of a medium is related to its relative permittivity (εr) and relative permeability (μr) by the formula:
n = sqrt(εr * μr)
We know that the speed of light in a vacuum (c) is related to the permittivity of free space (ε0) and the permeability of free space (μ0) by the formula:
c = 1/sqrt(ε0 * μ0)
We also know that the speed of light in a medium (v) is related to the speed of light in a vacuum (c) and the refractive index (n) by the formula:
v = c/n
We can combine these formulas to find an expression for the relative permittivity (εr):
εr = (n^2 * v^2) / (c^2 * μr)
Given that n = 1.5, μr = 5×10−7 H m−1, and c = 3×10^8 m/s, we can substitute these values into the formula to find εr:
εr = (1.5^2 * (3×10^8 m/s)^2) / ((3×10^8 m/s)^2 * 5×10−7 H m−1)
Solving this equation gives:
εr = 2.25 / 5×10−7
εr = 4.5×10^6
So, the relative permittivity of the medium is approximately 4.5×10^6.
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