Discussion
For plane monochromatic electromagnetic wave incident from a denser medium onto a rarer medium (refractive indices of the media are \(n_{1}\) and \(n_{2}\) respectively), the Brewster's angle and the critical angle are denoted by \(\theta_{B}\) and \(\theta_{C}\) respectively. Which of the following statements is correct ?
Answer: C
The correct statement is \(\theta_{B} \lt \theta_{C}\).
For light going from medium 1 (index \(n_{1}\)) into the rarer medium 2:
\[ \begin{aligned} \tan\theta_{B} &= n_{2}/n_{1} \\ \sin\theta_{C} &= n_{2}/n_{1} \end{aligned} \]
Both equal the same number less than 1. For any acute angle \(\tan\theta \gt \sin\theta\), so reaching the same value needs a smaller angle for the tangent: \(\theta_{B} \lt \theta_{C}\). This makes sense physically, because Brewster reflection needs a transmitted beam, which exists only below the critical angle.
The formulas \(\tan\theta_{B} = n_{1}/n_{2}\) and \(\sin\theta_{C} = n_{1}/n_{2}\) have the ratio upside down; the second is impossible since \(n_{1}/n_{2} \gt 1\). That also rules out \(\theta_{B} \gt \theta_{C}\).
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