Principle of Reversibility - Refraction of light - Param Himalaya
When a ray of light travels from medium 1 (say air) to the medium 2 (say water) along the path AOB, then refractive index of medium 2 w.r.t. medium 1 is given by
$${ }^{1}n_{2} = \frac{\sin i}{\sin r} \quad \dots(i)$$
When the path of the ray of light is reversed, it travels from medium 2 (water) to the medium 1 (air) along the path BOA,then the refractive index of medium 1 w.r.t. medium 2 is given by
$${ }^{2}n_{1} = \frac{\sin r}{\sin i} \quad \dots(ii)$$
Multiplying eqns. (i) and (ii), we get
$${ }^{1}n_{2} \times { }^{2}n_{1} = \frac{\sin i}{\sin r} \times \frac{\sin r}{\sin i} = 1$$
$${ }^{1}n_{2} = \frac{1}{{ }^{2}n_{1}}$$
Or
$${ }^{2}n_{1} = \frac{1}{{ }^{1}n_{2}}$$
Thus, refractive index of medium 2 w.r.t. medium 1 is reciprocal of the refractive index of medium 1 w.r.t. medium 2.