The refractive index of the core of an optical fiber is μ2 and that of the cladding is μ1. The angle of incidence on the face of the core so that the light ray just undergoes total internal reflection at the cladding is
A
sin−1(μ1μ2)
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B
sin−1(μ2−μ1)
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C
sin−1(√μ22−μ21)
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D
sin−1(√μ21−μ22)
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Solution
The correct option is Csin−1(√μ22−μ21) i=sin−1[√μ22−μ21]