A ray of light passes through four transparent media with...

Let $\theta_1, \theta_2, \theta_3$, and $\theta_4$ be the angles made by the light ray with the normal in the media with refractive indices $\mu_1, \mu_2, \mu_3$, and $\mu_4$ respectively.

Applying Snell's law at each parallel interface, we get:

$\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 = \mu_3 \sin \theta_3 = \mu_4 \sin \theta_4$

It is given that the incident ray AB in the first medium is parallel to the emergent ray CD in the fourth medium. This means the angle of the ray with the normal in the first medium is equal to the angle of the ray with the normal in the fourth medium.

Therefore, $\theta_1 = \theta_4$.

From the Snell's law equation, we have:

$\mu_1 \sin \theta_1 = \mu_4 \sin \theta_4$

Substituting $\theta_1 = \theta_4$, we get:

$\mu_1 \sin \theta_1 = \mu_4 \sin \theta_1$

$\Rightarrow \mu_1 = \mu_4$

Answer: $\mu_4=\mu_1$