The Brewster's angle for the air-glass interface is $(54.74)^{}^{\circ}$ . If a ray of light going from air to glass stikes at an angle of incidence $45^0$ , then the angle of refraction is
$\left(\tan (54.74)^0 = \sqrt{2}, \sin 45 = {1}/{\sqrt{2}}\right)$

Question

The Brewster's angle for the air-glass interface is

(54.74)^{}^{\circ}

. If a ray of light going from air to glass stikes at an angle of incidence

45^0

, then the angle of refraction is

\left(\tan (54.74)^0 = \sqrt{2}, \sin 45 = {1}/{\sqrt{2}}\right)

Quantrex Academy · Accepted Answer

Correct answer: B. ^ -1 ((1)/(2))

According to Brewster's law, the refractive index of the medium is given by

\mu = \tan \theta_B

.

Substituting the given Brewster's angle,

\mu = \tan(54.74^{}^{\circ}) = \sqrt{2}

.

Using Snell's law for refraction from air to glass,

1 \cdot \sin i = \mu \sin r

.

Substituting

i = 45^{}^{\circ}

and

\mu = \sqrt{2}

, we get

\sin 45^{}^{\circ} = \sqrt{2} \sin r

.

\dfrac{1}{\sqrt{2}} = \sqrt{2} \sin r

\sin r = \dfrac{1}{2}

r = \sin^{-1} \left(\dfrac{1}{2}\right)<br><br>Answer:

\sin^{-1} \left(\dfrac{1}{2}\right)

The Brewster's angle for the air-glass interface is (54.74)^ ^ . If...