Question

The Brewster's angle \( i_B \) for any interface should lie between

• A. \(30^\circ \text{ and } 45^\circ \)
• B. \(45^\circ \text{ and } 90^\circ \)
• C. \(0^\circ \text{ and } 30^\circ \)
• D. \(0^\circ \text{ and } 90^\circ \)
• E. \(30^\circ \text{ and } 60^\circ \)

Hint:

Brewster angle always lies between \(45^\circ\) and \(90^\circ\).

Answer 1

The Correct Option is B

Step 1: Understanding Brewster's Angle:
Brewster's angle (\(i_B\)), or the polarization angle, is a specific angle of incidence for light on a dielectric surface. At this angle, the reflected light is perfectly polarized with its electric field vector perpendicular to the plane of incidence.
Step 2: Key Formula - Brewster's Law:
Brewster's angle is defined by Brewster's Law, which relates the angle to the refractive indices of the two media:
\[ \tan(i_B) = n_{21} = \frac{n_2}{n_1} \] where \(n_1\) is the refractive index of the first medium (from which the light is incident) and \(n_2\) is the refractive index of the second medium.
Step 3: Detailed Explanation:
For most practical interfaces, light travels from a rarer medium (like air, with \(n_1 \approx 1\)) to a denser medium (like glass or water, where \(n_2>1\)).
In this common case, the relative refractive index \(n_{21} = \frac{n_2}{n_1}\) will be greater than 1.
\[ \frac{n_2}{n_1}>1 \] Therefore, according to Brewster's Law:
\[ \tan(i_B)>1 \] We know from trigonometry that \(\tan(45^\circ) = 1\), and the tangent function increases for angles between 0\(^\circ\) and 90\(^\circ\).
The condition \(\tan(i_B)>1\) implies that the angle \(i_B\) must be greater than 45\(^\circ\).
\[ i_B>45^\circ \] Since the angle of incidence cannot physically exceed 90\(^\circ\), the Brewster's angle must lie in the range:
\[ 45^\circ<i_B<90^\circ \] Step 4: Final Answer:
The Brewster's angle for any interface between two different optical media generally lies between 45\(^\circ\) and 90\(^\circ\).