Wave Optics Question 1

Question 1 - 24 January - Shift 1

Given below are two statements :

Statement I : If the Brewster’s angle for the light propagating from air to glass is $\theta_B$, then Brewster’s angle for the light propagating from glass to air is $\frac{\pi}{2}-\theta_B$.

Statement II : The Brewster’s angle for the light propagating from glass to air is $\tan ^{-1}(\mu_g)$ where $\mu_g$ is the refractive index of glass.

In the light of the above statements, choose the correct answer from the options given below :

(1) Both Statements I and Statement II are true.

(2) Statement I is true but Statement II is false.

(3) Both Statement I and Statement II are false.

(4) Statement I is false but Statement II is true.

Show Answer

Answer: (2)

Solution:

Formula: Laws of Refraction

$\mu_a \sin i_1=\mu_g \sin (90-i_1)$

$\tan i_1=\frac{\mu_g}{\mu_a}$

When going from glass to air

$\tan i_2=\frac{\mu_a}{\mu_g}=\cot i_1$

Hence

$i_2=\frac{\pi}{2}-i_1$