Question

What is Brewster's law? Derive the formula for Brewster's angle.

Hint:

The key to the derivation is the special condition: at Brewster's angle, the angle between the reflected and refracted rays is 90°. This simple geometric fact, when combined with Snell's law, leads directly to the \(\tan(i_p) = n\) formula.

Answer 1

Brewster's Law: This law states that for a particular angle of incidence, known as Brewster's angle or the polarizing angle (\(i_p\)), the reflected light from a transparent dielectric surface is completely plane-polarized. This occurs when the reflected ray and the refracted ray are perpendicular to each other.
Derivation:
Let unpolarized light be incident at Brewster's angle \(i_p\) on the surface separating two media (e.g., air and glass) with refractive indices \(n_1\) and \(n_2\) respectively. Let \(r\) be the angle of refraction.
According to Snell's law: \[ n_1 \sin(i_p) = n_2 \sin(r) \quad \cdots (1) \] By the condition of Brewster's law, the reflected and refracted rays are perpendicular. From the geometry of reflection and refraction, this means: \[ i_p + r = 90^{\circ} \] \[ r = 90^{\circ} - i_p \] Now, substitute this value of \(r\) into Snell's law (equation 1): \[ n_1 \sin(i_p) = n_2 \sin(90^{\circ} - i_p) \] Since \(\sin(90^{\circ} - \theta) = \cos(\theta)\), the equation becomes: \[ n_1 \sin(i_p) = n_2 \cos(i_p) \] Rearranging the terms to find the angle \(i_p\): \[ \frac{\sin(i_p)}{\cos(i_p)} = \frac{n_2}{n_1} \] \[ \tan(i_p) = \frac{n_2}{n_1} \] If the first medium is air or vacuum (\(n_1 \approx 1\)) and the second medium has refractive index \(n\) (\(n_2 = n\)), the formula simplifies to: \[ \tan(i_p) = n \] This is the formula for Brewster's angle.