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A transparent medium shows relation between \( i \) and \( r \) as shown. If the speed of light in vacuum
is \( c \) the Brewster angle for the medium is
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is \( c \) the Brewster angle for the medium is
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The correct answer is:
\( 60^{\circ} \)
(A)
By Brewster's law, \( \mathrm{n}=\tan \theta_{p} \)
Also \( n=\frac{\sin i}{\sin r} \)
But \( \frac{\sin i}{\sin r}=\frac{1}{\text { slope }} \) of given graph
\( \frac{\sin i}{\sin r}=\frac{1}{\tan 30^{\circ}}=\frac{1}{\frac{1}{\sqrt{3}}}=\sqrt{3} \)
\( \therefore \tan \theta_{p}=\mathrm{n}=\sqrt{3} \)
\( \theta_{p}=\tan ^{-1}(\sqrt{3})=60^{\circ} \)
By Brewster's law, \( \mathrm{n}=\tan \theta_{p} \)
Also \( n=\frac{\sin i}{\sin r} \)
But \( \frac{\sin i}{\sin r}=\frac{1}{\text { slope }} \) of given graph
\( \frac{\sin i}{\sin r}=\frac{1}{\tan 30^{\circ}}=\frac{1}{\frac{1}{\sqrt{3}}}=\sqrt{3} \)
\( \therefore \tan \theta_{p}=\mathrm{n}=\sqrt{3} \)
\( \theta_{p}=\tan ^{-1}(\sqrt{3})=60^{\circ} \)
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