Applying Snell’s law of refraction at A,

$$\mu = {{\sin 45^\circ } \over {\sin r}}$$

$ \Rightarrow $ r = $${\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right)$$ ....(1)

Applying the condition of total internal reflection at B,

i_c = $${\sin ^{ - 1}}\left( {{1 \over \mu }} \right)$$ ..........(2)

where i_c is the critical angle.

From figure, r + i_c = ${\pi \over 2}$

$ \Rightarrow $ $${\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right) = {\pi \over 2} - {\sin ^{ - 1}}\left( {{1 \over \mu }} \right)$$

$ \Rightarrow $ $${\sin ^{ - 1}}\left( {{1 \over {\sqrt 2 \mu }}} \right) = {\cos ^{ - 1}}\left( {{1 \over \mu }} \right)$$ = $${\sin ^{ - 1}}\left( {{{\sqrt {{\mu ^2} - 1} } \over \mu }} \right)$$

$ \Rightarrow $ $${{1 \over {\sqrt 2 \mu }} = {{\sqrt {{\mu ^2} - 1} } \over \mu }}$$

$ \Rightarrow $ $${\mu ^2} = 1 + {1 \over 2}$$

$ \Rightarrow $ $$\mu = \sqrt {{3 \over 2}} $$