A& ray of light passes four transparent glass slabs having&...

Applying Snell's law at B and C,

$\mu sini =$ constant or ${μ}_{1}{sini}_{B} = {μ}_{4}\text{sub}{i}_{c}$

But since $AB | | CD$

$∴ {i}_{B} = {i}_{C}$

or ${μ}_{1} = {μ}_{4}$

The Snell's law, which the students read in their plus two syllabus $\mu = \dfrac{sini}{sinr}$ comes from $\mu sini =$ constant

${μ}_{1} = {μ}_{1}{sini}_{1} = {μ}_{2}{sini}_{2}$

${μ}_{2} = \mu$ or $sini = \mu sinr$

or $\mu = \dfrac{sini}{sinr}$