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Question: A ray of light passes through four transparent media with refractive indices \(\mu _ { 1 } \cdot \mu...

A ray of light passes through four transparent media with refractive indices μ1⋅μ2,μ3\mu _ { 1 } \cdot \mu _ { 2 } , \mu _ { 3 }, and μ4\mu _ { 4 } as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

A

μ1=μ2\mu _ { 1 } = \mu _ { 2 }

B

μ2=μ3\mu _ { 2 } = \mu _ { 3 }

C

μ3=μ4\mu _ { 3 } = \mu _ { 4 }

D

μ4=μ1\mu _ { 4 } = \mu _ { 1 }

Answer

μ4=μ1\mu _ { 4 } = \mu _ { 1 }

Explanation

Solution

For successive refraction through difference media μsin⁡θ=\mu \sin \theta = constant.

Here as θ is same in the two extreme media. Hence μ1=μ4\mu _ { 1 } = \mu _ { 4 }