A) \[\sin \theta >\frac{8}{9}\]
B) \[\sin \theta <\frac{8}{9}\]
C) \[\sin \theta =\frac{1}{\sqrt{2}}\]
D) None of these
Correct Answer: A
Solution :
Angle of incidence at face \[AC\] is\[\theta \]. Since the ray suffers total internal reflection, \[\theta \] must be greater than critical. \[\Rightarrow \] \[\sin \theta >\sin C\] \[\sin \theta >\sin ({{\sin }^{-1}}\frac{1}{\mu })\] Now, for water-glass interface,\[\mu =\frac{{{\mu }_{g}}}{{{\mu }_{w}}}=\frac{9}{8}\]You need to login to perform this action.
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