Question

The relationship between brewster angle theta and the speed of light in the denser medium is

Answer 1

I know it was a MCQ in today's exam

we know that,

Refractive index, n = tanθ

c/v = tanθ

c = v.tanθ

Answer 2

$c = v tan(\theta_B)$

$v = c / tan( \theta_B)$

Step-by-step explanation:

Brewster's angle is defined as an angle at which an incident beam of polarized light is reflected after complete polarization. The incident light with an electric field parallel to the plane of incidence usually has a zero reflection coefficient at a particular angle between 0 and 90°.

Brewster's angle  is also called  polarizing angle.

The Brewster angle θB is defined by

$tan(\theta_B) = n_2/n_1$

if

  $n_1 = 1$

  $n_2 = n\\ tan(\theta_B) = n$

..............................equation (1)

where,

$n_1$= refractive index of the initial medium through which the light propagates (the "incident medium")

$n_2$ =  index of the other medium.

$\theta_B$ =  Brewster angle,

The refractive index formula is given as follows:

$n = c / v$    ..............................equation (2)

where,

$n$ = refractive index of the medium

$c$ = speed of light when in vacuum

$v$ = speed of light when in medium

from Equation(1) & (2)

$tan(\theta_B) = c / v\\c = v tan(\theta_B)\\v = c / tan(\theta_B)$