1.

Critical angle for certain medium is sin^(-1) (0.6). The polarizing angle of that medium is :

Answer»

<P>`ta^(-1) [1.5]`
`sin^(-1)[0.8]`
`tan^(-1)[1.6667]`
`tan^(-1)[0.6667]`

Solution :(c ) CRITICAL angle, `C = sin^(-1) (0.6)`
Sin (C ) = 0.6
`MU = (1)/(sinC)=((1)/(0.6))`
Polarising angle `i_(p) = tan^(-1) (mu) = tan^(-1) ((1)/(0.6))`
`tan^(-1) (1.6667)`


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