Question

The Brewster’s single for air to water interface is

• A. ${\tan }^{-1}(1.33)$
• B. ${\sin }^{-1}(1.33)$
• C. ${\cos }^{-1}(1.33)$
• D. ${\tan }^{-1}(\frac{1}{1.33})$
• E. ${\sin }^{-1}(\frac{1}{1.33})$

Answer 1

Answer: (A)

As we know, when partially reflected and refracted rays makes $90^{\circ }$ angle then both get polarised, i.e.
$i_p={\tan }^{-1}(\mu )$ (Brewester's law)
$\Rightarrow i_p={\tan }^{-1}\left(\frac{{\mu }_{\text{water }}}{{\mu }_{\text{air }}}\right)$
where, $i_p=$ Brewester's angle.
$\Rightarrow i_p={\tan }^{-1}\left(\frac{1.33}{1}\right)$
$i_p={\tan }^{-1}(1.33)$