1.

What will be the minimum angle of incidence such that the total internal reflection occurs on both the surfaces?

Answer»

Solution :Critical angle at `A=C_1`
`therefore sin C_1=mu_1/mu_2=sqrt2/2=1/sqrt2, C_1=45^@`
Critical angle at `B=C_2`
`therefore sinC_2=mu_3/mu_2=sqrt3/2 therefore C_2=60^@`
`therefore` Minimum angle of incidence for TOTAL internal REFLECTION to occur on both the slabs should be `60^@`
`therefore i_min=60^@`


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