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If light travels a distance $x$ in $t_1 \sec$ in air and $10 x$ distance in $t_2 \sec$ in a medium, the critical angle of the medium will be
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The correct answer is:
$\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$
From the formula $\sin C=\frac{1}{{ }_1 \mu_2} \Rightarrow \sin C={ }_2 \mu_1$
$=\frac{u_1}{u_2}=\frac{v_2}{v_1} \Rightarrow \sin C=\frac{10 x / t_2}{x / t_1}$
$\Rightarrow \sin C=\frac{10 t_1}{t_2} \Rightarrow C=\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$
$=\frac{u_1}{u_2}=\frac{v_2}{v_1} \Rightarrow \sin C=\frac{10 x / t_2}{x / t_1}$
$\Rightarrow \sin C=\frac{10 t_1}{t_2} \Rightarrow C=\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)$
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