Answer:
For glass-air interface, \[\sin {{i}_{c}}=\frac{1}{^{a}{{\mu }_{g}}}\] The critical angle \[i_{c}^{'}\] for glass-water interface is given by \[\sin i_{c}^{'}=\frac{1}{^{w}{{\mu }_{g}}}\] Now \[^{w}{{\mu }_{g}}{{<}^{a}}{{\mu }_{g}}\] \[\therefore \] \[\sin i_{c}^{'}>\sin {{i}_{c}}\] or \[i_{c}^{'}>{{i}_{c}}.\]
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