A light ray moving in medium - I (of refractive index n1) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index n2 of medium - II is decreased, then:
A
ray will move completely parallel to the interface
B
ray will be still totally internally reflected at interface
C
ray will be totally transmitted into medium-II only if angle of incidence is increased
D
ray will be totally transmitted in medium-II
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Solution
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As n2 decreases, iC=sin−1(n2n1) also decreases. So, condition i>ic is still satisfied and there will be still total internal reflection at interface. If angle of incidence is increases, then ray will be still totally internally reflected at interface because i>iC.
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