The Brewster's angle ib for an interface should be: (a) 30 ° < ib< 45 ° (b) 45 ° < ib< 90 ° (c) ib = 90 ° (d) 0 ° < ib< 30 °

Answer: (b). Brewster's law: tan ib = (n₂)/(n₁). For light going from rarer to denser medium (the usual case, e.g. air to glass), n₂ > n₁, so tan ib > 1. tan ib > 1 ⇒ 45° < ib < 90° ( ib = 90° would need n₂ → ∞). So the answer is (b).

The Brewster's angle ib for an interface should be: | NEET 2020 (Phase 1)

Q63NEET 2020 · PHASE 1Easy

Refraction, Glass Slab & Apparent Depth

The Brewster's angle ib for an interface should be:

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Answer: (b)

Brewster's law: $\tan i_b = \dfrac{n_2}{n_1}$ .
For light going from rarer to denser medium (the usual case, e.g. air to glass), $n_2 > n_1$ , so $\tan i_b > 1$ .
$\tan i_b > 1 \Rightarrow 45^\circ < i_b < 90^\circ$ ( $i_b = 90^\circ$ would need $n_2 \to \infty$ ).
So the answer is (b).

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