Q169 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
The Brewster's angle $\theta_B$ for air-glass interface should be such that
(refractive index of glass $= \dfrac{3}{2}$)
(refractive index of glass $= \dfrac{3}{2}$)
- A$\theta_B = 90^\circ$
- Correct answerB$45^\circ < \theta_B < 90^\circ$
- C$30^\circ < \theta_B < 45^\circ$
- D$0^\circ < \theta_B < 45^\circ$
Solution
According to Brewster's law, the tangent of the polarizing angle (Brewster's angle) is equal to the refractive index of the medium.
$\tan \theta_B = \mu$
Given $\mu = \dfrac{3}{2} = 1.5$
$\tan \theta_B = 1.5$
We know that $\tan 45^\circ = 1$ and $\tan 90^\circ \rightarrow \infty$.
Since $1 < 1.5 < \infty$, the angle $\theta_B$ must lie between $45^\circ$ and $90^\circ$.
$45^\circ < \theta_B < 90^\circ$
Answer: $45^\circ < \theta_B < 90^\circ$
$\tan \theta_B = \mu$
Given $\mu = \dfrac{3}{2} = 1.5$
$\tan \theta_B = 1.5$
We know that $\tan 45^\circ = 1$ and $\tan 90^\circ \rightarrow \infty$.
Since $1 < 1.5 < \infty$, the angle $\theta_B$ must lie between $45^\circ$ and $90^\circ$.
$45^\circ < \theta_B < 90^\circ$
Answer: $45^\circ < \theta_B < 90^\circ$
Hence, the correct answer is option: 'B'.