1. 0^° < i_b < 30^° 2. 30^° < i_b < 45^° 3. 45^° < i_b < 90^° 4. i_b = 90^°

Answer: 3. 45^° 1 for an ordinary interface, tan i_b>1 , so i_b>45^° . Also tan i_b is finite, so i_b<90^° . Hence 45^°

The Brewsters angle i_b for an interface should be :

1. \(0^\circ < i_b < 30^\circ\)

2. \(30^\circ < i_b < 45^\circ\)

3. \(45^\circ < i_b < 90^\circ\)

4. \(i_b = 90^\circ\)

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

Best answer

Answer: 3. \(45^\circ < i_b < 90^\circ\)

Brewster's law: \(\tan i_b=\mu\), where \(\mu\) is the relative refractive index of the interface.

Since \(\mu>1\) for an ordinary interface, \(\tan i_b>1\), so \(i_b>45^\circ\). Also \(\tan i_b\) is finite, so \(i_b<90^\circ\).

Hence \(45^\circ

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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