A ray of light passes through four transparen | vedclass

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(A) According to Snell's law for a series of parallel interfaces,the product of the refractive index and the sine of the angle of incidence at each in

Vedclass · Accepted Answer

$n_{1} = n_{4}$

Vedclass · Answer

(A) According to Snell's law for a series of parallel interfaces,the product of the refractive index and the sine of the angle of incidence at each interface remains constant.Let $	heta_{1}$ be the angle of incidence in the first medium and $	heta_{4}$ be the angle of emergence in the fourth medium.Applying Snell's law at each interface:$n_{1} \sin 	heta_{1} = n_{2} \sin 	heta_{2} = n_{3} \sin 	heta_{3} = n_{4} \sin 	heta_{4}$Given that the emergent ray $DE$ is parallel to the incident ray $AB$,the angle of incidence $	heta_{1}$ must be equal to the angle of emergence $	heta_{4}$ (i.e.,$	heta_{1} = 	heta_{4}$).Therefore,$n_{1} \sin 	heta_{1} = n_{4} \sin 	heta_{1}$.Since $\sin 	heta_{1} 
eq 0$,we get $n_{1} = n_{4}$.

$A$ ray of light passes through four transparent media with refractive indices $n_{1}$,$n_{2}$,$n_{3}$,and $n_{4}$ as shown. The surfaces of all media are parallel. If the emergent ray $DE$ is parallel to the incident ray $AB$,then

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