Question

The critical angle for glass-water boundary is C. If $ {{\mu }_{g}}=1.5 $ 1.5 and $ {{\mu }_{\omega }}=\frac{4}{3} $ then the value of C is:

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

Answer 1

Answer: (C)

From the relation $ \sin C=\frac{{{\mu }_{w}}}{{{\mu }_{g}}}=\frac{4/3}{3/2}=\frac{8}{9} $ $ C={{\sin }^{-1}}\frac{8}{9} $