Three right angled prisms of refractive indices $\mu_1$, $\mu_2$ and $\mu_3$ are joined together ao that the faces of the middle prism are each in contact with one of the outside prisms. If the ray passes through the composite block undeviated, then 

$\alpha = 90-r_i$

$\beta = 90-r_2$

$\gamma = 90-r_3$

$\begin{array}{l} Applying\, snell's\, Law\, at\, each\, surface \\ AtB, \\ \sin i={ \mu _{ 1 } }\sin { r_{ 1 } } \Rightarrow { \sin ^{ 2 } }i=\mu _{ 1 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 1 } } \\ At\, C \\ { \mu _{ 1 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 2 } }\sin { r_{ 2 } } \Rightarrow \mu _{ 1 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 2 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 2 } } \\ At\, D \\ { \mu _{ 2 } }\sin \left( { 90-{ r_{ 2 } } } \right) ={ \mu _{ 3 } }\sin { r_{ 3 } } \Rightarrow \mu _{ 2 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 2 } }=\mu _{ 3 }^{ 2 }{ \sin ^{ 2 } }{ r_{ 3 } } \\ At\, E, \\ { n_{ 3 } }\sin \left( { 90-{ r_{ 3 } } } \right) =\sin \left( { 90-i } \right) \Rightarrow \mu _{ 3 }^{ 2 }{ \cos ^{ 2 } }{ r_{ 3 } }={ \cos ^{ 2 } }i \\ Adding\, all\, above\, equations\, \, gives, \\ 1+\mu _{ 2 }^{ 2 }=\mu _{ 1 }^{ 2 }+\mu _{ 3 }^{ 2 } \\ Hence,\, the\, option\, \, A\, is\, the\, correct\, answer. \end{array}$