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