The critical angle of a certain medium is \({\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\). The polarizing angle of the medium is

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Question

The critical angle of a certain medium is \({\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\). The polarizing angle of the medium is

\({\sin ^{ - 1}}\left( {\frac{4}{5}} \right)\)
\({\tan ^{ - 1}}\left( {\frac{5}{3}} \right)\)
\({\tan ^{ - 1}}\left( {\frac{3}{4}} \right)\)
\({\tan ^{ - 1}}\left( {\frac{4}{3}} \right)\)

Answer 1

Concept:

Brewster angle

The angle of incidence at which a beam of unpolarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarising or Brewster angle. It is denoted by ip.

The British Physicist David Brewster found the relationship between Brewster angle (ip) and refractive index (μ) –

μ = tan ip

This relation is known as Brewster law.

Calculation:

Given:

Critical angle,

\({i_c} = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\;\;\;\;\therefore \sin {i_c} = \frac{3}{5}\)

As \(\mu = \frac{1}{{\sin {i_c}}} = \frac{5}{3}\)

According to Brewster’s law

tan i_p = μ

Where i_p is the polarising angle

∴ \(\tan {i_p} = \frac{5}{3}\;{i_p} = {\tan ^{ - 1}}\left( {\frac{5}{3}} \right)\)

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The critical angle of a certain medium is \({\sin ^{ - 1}}\left( {\frac{3}{5}} \right)\). The polarizing angle of the medium is

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