Question 237: Waves Optics

Question

The polarising angle of transparent medium is ' $	heta$ '. Let the speed of light in the medium be ' v '. Then the relation between ' $	heta$ ' and ' $\mathbf{v}$ ' is [ $\mathrm{c}=$ velocity of light in air]

Accepted Answer

Step 1: BrewsterÃ¢ÂÂs Law
The polarising angle (Brewster's angle) $	heta_p$ is defined by:
$ 	an 	heta_p = \mu $
where $\mu$ is the refractive index of the medium with respect to air.
Step 2: Relation between refractive index and velocity
$ \mu = \frac{c}{v} $
where $c$ = speed of light in vacuum (or air approximately),
$v$ = speed of light in the medium.
Step 3: Substitute into BrewsterÃ¢ÂÂs Law
$ 	an 	heta = \frac{c}{v} $
$ 	heta = 	an^{-1}\left(\frac{c}{v}ight) $
Step 4: Compare options
We want an expression involving $\frac{v}{c}$. Note:
$ 	heta = 	an^{-1}\left(\frac{c}{v}ight) = \cot^{-1}\left(\frac{v}{c}ight) $
Ã¢ÂÂ Correct option:
Option C
$ 	heta = \cot^{-1}\left(\frac{v}{c}ight) $

Answer

$\quad 	heta=\sin ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}ight)$

Answer

$	heta=	an ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}ight)$

Answer

$	heta=\cot ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}ight)$

Answer

$\quad 	heta=\cos ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}ight)$

Waves Optics

The polarising angle of transparent medium is ' \theta '. Let the speed of light in the medium be ' v '. Then the relation between ' \theta ' and ' \mathbf{v} ' is [ \mathrm{c}= velocity of light in air]

Choose the correct answer: