Answer to Question #164101 in Optics for wilder

Question

 Show that, when the condition for completely polarized light is satisfied, the reflected and refracted

   beams are at right angles to one another.

     

Answer 1

When a beam of light strikes the surface of a piece of transparent material, the beam is split, part of the beam is refracted and the other part of the beam is reflected. From the demonstration, the angle of reflection and the angle of incidence. Moreover, for one particular angle of incidence called the Brewster's angle $\theta _B$ , the reflected beam is completely polarized. At Brewster's angle, the angle between the refracted and reflected light is $90^{\circ }$ . Brewster's angle can be related to the index of refraction of the reflecting surface by using Snell's law: $sin\theta _1n_1=sin\theta _2n_2$ . Substitution of:

$\theta _1=\theta _B$, $n_1=n_{air}=1\:$ and $n_2=n\:$ into the equation of Snell's law we get:

$sin\theta _B=nsin\theta _2$

since $sin\theta _2=sin\left(90^{\circ }-\theta _B\right)$ the expression for n is written as: $\:n=\frac{\:\:sin\theta \:_B}{\:cos\theta \:_B}=tan\theta \:_B$ .