alleryBrewster's law: angle of polarization i_B from glass-air interface satisfies:
Acos i_B = n
Btan i_B = n (refractive index)
Csin i_B = n
Di_B = 45°
Answer & Solution
Correct answer: B. tan i_B = n (refractive index)
At Brewster's angle, reflected light is completely polarized (E perpendicular to plane of incidence). tan i_B = n.
For glass (n = 1.5): i_B ≈ 56.3°.
Reflected and refracted rays are perpendicular at this angle.
Keep practising Wave Optics
Browse all Wave Optics questions for JEE Advanced, or explore more Physics topics.
Related questions
Polarisation of light is direct experimental evidence that light is a:The resolving power (RP) of a diffraction grating with N slits in the m-th order is:In Young's double-slit experiment, the fringe width β is given by:In Young's experiment, a BRIGHT fringe (constructive interference) requires path differencIn single-slit diffraction, the central maximum is:BREWSTER'S LAW for the polarising angle θ_B relates to the refractive index n as:HUYGENS' PRINCIPLE states that every point on a wavefront acts as a:Light passes from air to glass ($n = 1.5$). Brewster's angle is: