A) (i) and (iii)
B) (i) and (iv)
C) (ii) and (iii)
D) (ii) and (iv)
Correct Answer: D
Solution :
We know \[C={{\sin }^{-1}}\left( \frac{1}{\mu } \right)\] Given critical angle \[{{i}_{B}}>{{i}_{A}}\] So \[{{\mu }_{B}}<{{\mu }_{A}}\]i.e. B is rarer and A is denser. Hence light can be totally internally reflected when it passes from A to B Now critical angle for A to B \[{{C}_{AB}}={{\sin }^{-1}}\left( \frac{1}{_{B}{{\mu }_{A}}} \right)\]\[={{\sin }^{-1}}\left[ _{A}{{\mu }_{B}} \right]\] \[={{\sin }^{-1}}\left[ \frac{{{\mu }_{B}}}{{{\mu }_{A}}} \right]\]\[={{\sin }^{-1}}\left[ \frac{\sin {{i}_{A}}}{\sin {{i}_{B}}} \right]\]You need to login to perform this action.
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