Light traveling through three transparent substances follows the path shown in Fig. Arrange the indices of refraction in order from smallest to largest. Note that total internal reflection does occur on the bottom surface of medium 2
A
n1<n2<n3
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B
n2<n1<n3
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C
n1<n3<n2
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D
n3<n1<n2
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Solution
The correct option is Bn3<n1<n2 As the ray moves toward the normal while entering medium 2 from 1, we have n2>n1 For total internal reflection at interface of 2 and 3,n2>n3. Besides n3 should also be less than n1 or else ray would have emerged in medium 3, parallel to its path in medium 1. Hence, n3<n1<n2 is the correct order.