Q167 / 170 · Manipal (MET) 2020 · Physics

The critical angle for glass water interface $\left(\right.$ if $\left._a \mu_g=\frac{3}{2},{ }_a \mu_w=\frac{4}{3}\right)$

  • Correct answer
    A
    $\sin ^{-1}\left(\frac{8}{9}\right)$
  • B
    $\sin ^{-1}\left(\frac{9}{8}\right)$
  • C
    $\sin ^{-1}\left(\frac{3}{2}\right)$
  • D
    $\sin ^{-1}\left(\frac{4}{3}\right)$

Solution

Critical angle $\left(\theta_c\right)$ for glass water interface can be written as $\operatorname{Sin} \theta_c=\frac{{ }_a \mu_w}{{ }_a \mu_g}=\frac{4 / 3}{3 / 2}=\frac{8}{9}$ $\Rightarrow \quad \theta_c=\sin ^{-1}(8 / 9)$

Hence, the correct answer is option: 'A'.

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