Critical angle uses \(\frac{\text{rarer}}{\text{denser}}\) and must always be \(< 1\).
To determine the critical angle for the glass-water interface, we need to use Snell's Law and the concept of critical angle. The critical angle is the angle of incidence for which the angle of refraction is 90 degrees.
Snell's Law is given by:
\(\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2\)
For the critical angle, \(\theta_2 = 90^\circ\), hence \(\sin \theta_2 = 1\). So, the formula becomes:
\(\mu_g \sin \theta_c = \mu_w \sin 90^\circ\)
Substituting, \(\sin 90^\circ = 1\), we get:
\(\mu_g \sin \theta_c = \mu_w\)
This can be rearranged as:
\(\sin \theta_c = \frac{\mu_w}{\mu_g}\)
Given \(\mu_g = \frac{3}{2}\) and \(\mu_w = \frac{4}{3}\), substituting these values gives:
\(\sin \theta_c = \frac{\frac{4}{3}}{\frac{3}{2}} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}\)
Thus, the critical angle \(\theta_c\) is:
\(\theta_c = \sin^{-1}\left(\frac{8}{9}\right)\)
Therefore, the correct answer is:
\( \sin^{-1}\left(\frac{8}{9}\right) \)