> Q. 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 – LIVE ANSWER TODAY

Q. 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

In This post you will get answer of Q.
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:

As $n_2$ decreases $, i_c = sin^{-1} left( frac{n_2}{n_1} right) $ 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 $i > i_c$

Leave a Comment