A light ray moving in medium - I (of refractive index $n_{1}$ ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index $n_{2}$ of medium - II is decreased, then:

ray will move completely parallel to the interface

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ray will be still totally internally reflected at interface

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ray will be totally transmitted into medium-II only if angle of incidence is increased

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ray will be totally transmitted in medium-II

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Solution

The correct option is C ray will be still totally internally reflected at interface
As $n_{2}$ decreases, $i_{C} = {sin}^{- 1} (\frac{n_{2}}{n_{1}})$ also decreases.
So, condition $i > i_{c}$ 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 $i > i_{C}$ .

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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 light ray moving in medium - I (of refractive index $n_{1}$ ) is incident on interface of two media and it is totally internally reflected at the interface. Now, refractive index $n_{2}$ of medium - II is decreased, then: