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

Question

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

MARKS Content Team · Accepted Answer

Correct answer: (B) ray will be still totally internally reflected at interface. As n_2 decreases, i_C=sin^(-1)⁡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 increased, then ray will be still totally internally reflected at interface because i>i_C.

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

Solution

More Ray Optics Questions

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

Solution

More Ray Optics Questions

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