Question

Unpolarised light is incident from air on a plane surface of a material of refractive index ‘μ’. At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation ?

• A. Reflected light is polarised with its electric vector parallel to the plane of incidence
• B. \(i=sin^{-1}\left(\frac{1}{\mu}\right)\)
• C. Reflected light is polarised with its electric vector perpendicular to the plane of incidence
• D. \(i=tan^{-1}\left(\frac{1}{\mu}\right)\)

Answer 1

The Correct Option is C

The problem involves the concept of light polarization and Brewster's angle. When unpolarised light is incident on a surface, certain conditions can result in the reflected light being polarized.

Let's go through the steps to understand why the correct option is "Reflected light is polarised with its electric vector perpendicular to the plane of incidence".

1. When unpolarized light is incident at a specific angle, known as Brewster's angle or the polarizing angle, the reflected light becomes completely polarized.
2. According to Brewster's Law, the angle at which light is perfectly polarized upon reflection is given by: \( \tan(i) = \mu \) where \(i\) is the angle of incidence and \(\mu\) is the refractive index of the material.
3. Due to the properties of polarization, the reflected light at Brewster's angle is polarized with its electric vector perpendicular to the plane of incidence. This attribute is due to the nature of electromagnetic waves and their interaction with surfaces.
4. Given the condition that the reflected and refracted rays are perpendicular, it aligns with Brewster's condition, validating that the reflected light will indeed have its electric vector perpendicular to the plane of incidence.

Let's evaluate the options:

• Reflected light is polarised with its electric vector parallel to the plane of incidence: This is incorrect because, according to Brewster's law, the electric vector is actually perpendicular to the plane of incidence.
• \(i = \sin^{-1}\left(\frac{1}{\mu}\right)\): This relation doesn't describe Brewster's angle. It is incorrect since the correct relation should use the tangent, not sine.
• Reflected light is polarised with its electric vector perpendicular to the plane of incidence: This is the correct option based on the explanation provided above.
• \(i = \tan^{-1}\left(\frac{1}{\mu}\right)\): This is incorrect since based on Brewster's Law, it should be \(i = \tan^{-1}(\mu)\).

Conclusively, the correct answer is that the reflected light is polarized with its electric vector perpendicular to the plane of incidence.