The refractive indices of four media, whose surfaces are all parallel, are n₁, n₂, n₃ and n₄ as shown in the figure. AB is the incident ray and DE the emergent ray. If DE is parallel to AB, then .

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(4) n₄ = n₁. Given: four media with parallel boundaries; AB enters at the top and DE leaves at the bottom, parallel to AB. Idea: write Snell's law at each boundary in turn and see what survives. n₁ sin θ₁ = n₂ sin θ₂ at the first surface, n₂ sin θ₂ = n₃ sin θ₃ at the second, and n₃ sin θ₃ = n₄ sin θ₄ at the third. Because the surfaces are parallel, the angle a ray makes with the normal is the same where it leaves a layer as where it entered, so the three equations chain into n₁ sin θ₁ = n₄ sin θ₄. DE parallel to AB means θ₄ = θ₁, so sin θ₄ = sin θ₁ and therefore n₄ = n₁. The middle indices drop out completely: a stack of parallel slabs shifts a ray sideways, but only the first and last media decide the direction it ends up travelling in. So the condition is n₄ = n₁.

The refractive indices of four media, whose surfaces are all parallel, are n₁, n₂, n₃ and n₄ as shown in the figure. AB is the incident ray and DE the emergent ray. If DE is parallel to AB, then ______.

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