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(D) When a light ray passes through a series of parallel media and finally emerges into the original medium,the product of the relative refractive ind

Vedclass · Accepted Answer

$_{w}n_{g} = \frac{_{a}n_{g}}{_{a}n_{w}}$

Vedclass · Answer

(D) When a light ray passes through a series of parallel media and finally emerges into the original medium,the product of the relative refractive indices is equal to $1$. For the given sequence (air $ightarrow$ water $ightarrow$ glass $ightarrow$ air),the principle of reversibility and the property of parallel slabs imply: $_{a}n_{w} 	imes _{w}n_{g} 	imes _{g}n_{a} = 1$. We know that $_{g}n_{a} = \frac{1}{_{a}n_{g}}$. Substituting this into the equation: $_{a}n_{w} 	imes _{w}n_{g} 	imes \frac{1}{_{a}n_{g}} = 1$. Rearranging the terms to solve for $_{w}n_{g}$: $_{w}n_{g} = \frac{_{a}n_{g}}{_{a}n_{w}}$. Thus,option $D$ is the correct relation.

$A$ light ray undergoes refraction from air to water,water to glass,and finally from glass back to air. If the emergent ray is parallel to the incident ray,which of the following relations is correct? ($n_{a}, n_{w}, n_{g}$ represent the refractive indices of air,water,and glass respectively.)

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