Question 384: Waves Optics

Question

The Brewster's angle for the glass-air interface is $(54.74)^{\circ}$. If a ray of light passing from air to glass strickes at an angle of incidence $45^{\circ}$, then the angle of refraction is
$$\left[	an (54.74)^{\circ}=\sqrt{2}, \sin 45=\frac{1}{\sqrt{2}}ight]$$

Accepted Answer

According to Brewster's law, the refractive index of glass $\mu_2$ can be determined using the Brewster's angle:
$ \mu_2 = 	an(i_p) = 	an(54.74^{\circ}) = \sqrt{2} $
Using Snell's law, we relate the angles and refractive indices:
$ \mu_1 \sin i = \mu_2 \sin r $
Given that $\mu_1 = 1$ (for air), $i = 45^{\circ}$, and $\mu_2 = \sqrt{2}$:
$ 1 	imes \sin(45^{\circ}) = \sqrt{2} \sin r $
Substituting for $\sin(45^{\circ})$:
$ \frac{1}{\sqrt{2}} = \sqrt{2} \sin r $
Simplifying:
$ \frac{1}{\sqrt{2} 	imes \sqrt{2}} = \sin r $
$ \sin r = 0.5 $
Thus, the angle of refraction $r$ is:
$ r = \sin^{-1}(0.5) $

Answer

$\sin ^{-1}(\sqrt{2})$

Answer

$\sin ^{-1}(1)$

Answer

$\sin ^{-1}(05)$

Answer

$\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Waves Optics

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

Choose the correct answer: