Question

When visible light moves from glass refractive index 1.6 to air refractive index 1.0. What is the polarizing angle (Brewster angle)? ​

Answer 1

Answer:

$polrizing \: angle \: \: \: p = {58}^{0}$

Explanation:

$n \: \: of \: \: glass \: \: n2 = 1.6 \\ n \: \: of \: air \: \: \: n1 = 1.0 \\ according \: brewsters \: law \\ \: \: \: \: \: \tan(p) \: = \frac{n2}{n1}$

$tan \: p \: = \frac{1.6}{1.0} \: \: \\ \tan(p) = 1.6 \\ p = { \tan }^{ - 1} 1.6$

$p = {58}^{0}$

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