1. Tardigrade
  2. Question
  3. Physics
  4. Given below are two statements: Statement I: If the Brewster's angle for the light propagating from air to glass is θ B , then the Brewster's angle for the light propagating from glass to air is (π/2)-θm Statement II: The Brewster's angle for the light propagating from glass to air is tan -1(μ g ) where μ g is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below:

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Q. Given below are two statements :
Statement I : If the Brewster's angle for the light propagating from air to glass is $\theta_{ B }$, then the Brewster's angle for the light propagating from glass to air is $\frac{\pi}{2}-\theta_m$
Statement II : The Brewster's angle for the light propagating from glass to air is $\tan ^{-1}\left(\mu_{ g }\right)$ 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:

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Solution:

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$ \mu_{ a } \sin i _1=\mu_{ g } \sin \left(90- i _1\right)$
$ \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
$i _2=\frac{\pi}{2}- i _1$