A transparent solid cylindrical rod has a refractive index of 2√3. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle θ for which the light ray grazes along the wall of the rod is -
sin−1(12)
sin−1(2√3)
sin−1(1√3).
sin−1(√32)
A
sin−1(12)
B
sin−1(2√3)
C
sin−1(√32)
D
sin−1(1√3).
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Solution
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I×sin90=2√3sin(90−α) ⇒cosα=√32 So, sinα=√1−34=12 Now, 1×sinθ=2√3sinα=2√3×12=1√3 ⇒θ=sin−11√3
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A transparent solid cylindrical rod has a refractive index of 2√3. It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure.
The incident angle θ for which the light ray grazes along the wall of the rod is: