For light incident from air onto a transparent dielectric surface, the Brewster's angle $i_b$ is determined by the refractive index $n$ of the dielectric medium according to $\tan(i_b) = n$. Considering typical transparent dielectric materials, which generally have $n > 1$, what is the characteristic range for $i_b$?
$45^\circ < i_b < 90^\circ$
To determine the characteristic range for Brewster's angle \(i_b\) when light is incident from air onto a transparent dielectric surface, we utilize Brewster's law. Brewster's law states that the Brewster's angle is given by:
\(\tan(i_b) = n\)
where \(n\) is the refractive index of the dielectric medium. For typical transparent dielectric materials, the refractive index \(n\) is greater than 1.
To find the range of \(i_b\), consider the behavior of the tangent function and the values of \(n\):
- The tangent function, \(\tan(\theta)\), increases from 0 to infinity as the angle \(\theta\) increases from \(0^\circ\) to \(90^\circ\).
- For \(n > 1\), the equation \(\tan(i_b) = n\) implies that \(i_b\) must be an angle where the tangent gives a value greater than 1.
The tangent of \(45^\circ\) is exactly 1, which means Brewster's angle must be greater than \(45^\circ\) to satisfy \(\tan(i_b) = n > 1\). As such, the Brewster's angle \(i_b\) must be within the range:
\(45^\circ < i_b < 90^\circ\)
Therefore, the correct answer is the option: \(45^\circ < i_b < 90^\circ\).
This range is due to the constraints imposed by the value of the tangent function and the requirement that the refractive index \(n\) is greater than 1, which is commonly the case for dielectric materials.
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