Material A has a critical angle {i_A}, and ma | vedclass

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(D) We know that the critical angle $C$ is given by $C = {\sin ^{ - 1}}\left( {\frac{1}{\mu }} \right)$.Given that the critical angle ${i_B} > {i_A}$,

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$(ii)$ and $(iv)$

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(D) We know that the critical angle $C$ is given by $C = {\sin ^{ - 1}}\left( {\frac{1}{\mu }} \right)$.Given that the critical angle ${i_B} > {i_A}$,it implies that the refractive index ${\mu _B} This means material $B$ is optically rarer and material $A$ is optically denser.Total internal reflection occurs when light travels from a denser medium to a rarer medium. Therefore,light can be totally internally reflected when it passes from $A$ to $B$. Thus,statement $(ii)$ is correct.Now,the critical angle ${C_{AB}}$ for the interface between $A$ and $B$ is given by ${C_{AB}} = {\sin ^{ - 1}}\left( {\frac{{{\mu _B}}}{{{\mu _A}}}} \right)$.Since $\sin {i_A} = \frac{1}{{{\mu _A}}}$ and $\sin {i_B} = \frac{1}{{{\mu _B}}}$,we have ${\mu _A} = \frac{1}{{\sin {i_A}}}$ and ${\mu _B} = \frac{1}{{\sin {i_B}}}$.Substituting these values,we get ${C_{AB}} = {\sin ^{ - 1}}\left( {\frac{{1/\sin {i_B}}}{{1/\sin {i_A}}}} \right) = {\sin ^{ - 1}}\left( {\frac{{\sin {i_A}}}{{\sin {i_B}}}} \right)$. Thus,statement $(iv)$ is correct.

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