​ State the principle of reversibility of light. Hence show that aµb = 1/bμa ​

Question

State the principle of reversibility of light. Hence show that ^aµ_b = \(\frac{1}{^b\mu_a}\)

Answer 1

Principle: When final path of a ray of light after any number of reflections and refractions is reversed, the ray retraces back its entire path.

In Fig. ray of light travelling along AO in medium a is refracted along OB in medium b, at the interface XY.

Let ∠AON = i and ∠BON' = r

For incident ray

From Snell's law of refraction,

\(\frac{sin\ i}{sin\ r}\) = ^aµ_b ............(1)

Let a plane mirror M be held ⊥ to OB. The ray retraces its path and emerges along OA.

For the reversed beam

Angle of incidence i = angle of refraction r.

According to Snell's law,

\(\frac{sin\ r}{sin\ i}\) = ^bµ_a ..........(2)

Multiplying (1) and (2),

\(\frac{sin\ i}{sin\ r} \times \frac{sin\ r}{sin\ i}\) = ^aµ_b x ^bµ_a

i.e. 1 = ^aµ_b x ^bµ_a
or ^aµ_b = \(\frac{1}{b\mu_a}\)