Question

Discuss the refraction of light through a compound slab and show that ^aµ_g = ^aµ_w x _wµ_g.

Answer 1

Let us consider a compound slab made of two transparent media say water (w) and glass (g) bounded by parallel faces as shown in Fig. Let this compound slab be held in air (a).

A ray AO incident on the slab at O making an angle of incidence i and is refracted at angle r₁ along OB. At point B, it is again refracted at angle r₂ and finally refracted at C making an angle of refraction i at C because the medium on both sides of the slab is the same. So

At point O

From Snell's law, we have

\(\frac{sin\ i}{\sin\ r_1}\) = ^aµ_w ............(1)

At point B

\(\frac{sin\ r_1}{\sin\ r_2}\) = ^wµ_g ..........(2)

AT point C

From Snell's law, we have

\(\frac{sin\ r_2}{\sin\ i}\) = ^gµ_a .........(3)

Multiplying eqns. (1), (2) and (3), we get

\(\frac{sin\ i}{\sin\ r_1} \times \frac{sin\ r_1}{\sin\ r_2} \times\) \(\frac{sin\ r_2}{\sin\ i}\) = ^aµ_w x ^wµ_g x ^gµ_a

or 1 = ^aµ_w x ^wµ_g x ^gµ_a
or ^aµ_w x ^wµ_g = \(\frac{1}{^g\mu_a}\)

or ^aµ_w x ^wµ_g = ^aµ_g [∵ ^aµ_g = \(\frac{1}{^g\mu_a}\)]

or ^aµ_g = ^aµ_w x ^wµ_g.