A ray of light falls on a transparent glass slab of refractive index 1.62 . If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

Question

A ray of light falls on a transparent glass slab of refractive index 1.62 . If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is (A)  tan-1 (1.62)  (B)  tan-1 ((1/1.62))  (C)  tan-1 (1.33)  (D)  tan-1 ((1/1.33))

Tardigrade Admin · Accepted Answer

Brewster discovered that when a beam of unpolarised light is reflected from a transparent medium (refractive index

= μ)

, the reflected light is completely plane polarised at a certain angle of incidence (called the angle of polarisation

θ_{p}

)
Also,

μ = tan θ_{p}

∴

Here,

θ_{p} = θ_{i} = tan^{- 1} (1.62)

Q. A ray of light falls on a transparent glass slab of refractive index $1.62$ . If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

$t a n^{- 1} (1.62)$

$t a n^{- 1} (\frac{1}{1.62})$

$t a n^{- 1} (1.33)$

$t a n^{- 1} (\frac{1}{1.33})$

Q. A ray of light falls on a transparent glass slab of refractive index 1.62 . If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

tan−1(1.62)

tan−1(1.621​)

tan−1(1.33)

tan−1(1.331​)

Q. A ray of light falls on a transparent glass slab of refractive index $1.62$ . If the reflected ray and the refracted ray are mutually perpendicular, the angle of incidence is

$t a n^{- 1} (1.62)$

$t a n^{- 1} (\frac{1}{1.62})$

$t a n^{- 1} (1.33)$

$t a n^{- 1} (\frac{1}{1.33})$