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μ=sin450sinr⇒sinr=1μ√2..........(1)

From figure,

i1=90−r∴sin(900−r)=1μNow,⇒cosr=1μ..........(2)cosr=√1−sin2r=√1−12μ2=√2μ2−12μ2...........(3)fromequation(2)and(3)1μ=√2μ2−12μ2

Squaring both sides then solving we get =√32

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