Question

Unpolarized 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 polarized with its electric vector parallel to the plane of incidence
(B) Reflected light is polarized with its electric vector perpendicular to the plane of incidence
(C) \[i={{\sin }^{-1}}(1/\mu )\]
(D) \[i={{\tan }^{-1}}(1/\mu )\]

Answer 1

Hint: We are given a problem in which an unpolarized light is incident on a plane surface of a given refractive index. The angle of incidence, the angle of reflection & angle of refraction all are made with the normal to the surface. This problem involves the use of Brewster law.

Complete step by step answer:
According to Brewster law when an unpolarized light is incident on a reflecting surface, then at some particular angle of incidence some of the light is reflected back while the other gets refracted. The reflected light is completely polarized and the angle between the reflected ray and the refracted ray is \[{{90}^{0}}\]. When reflected light and refracted light are perpendicular, reflected light is polarized with electric field vectors perpendicular to the plane of incidence.

So, the correct answer is “Option B”.

Note:
When the angle of incidence equals to Brewster angle then the Brewster law is \[\mu =\tan ({{i}_{p}})\]
The reflected light is linearly polarized with its electric field vectors perpendicular to the plane of incidence and parallel to the plane of the surface from which it is reflecting