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A ray of light passes through four transparent media with refractive indices $\mu_1, \mu_2 \mu_3$, and $\mu_4$ as shown in the figure. The surfaces of all media are parallel. If the emergent ray $C D$ is parallel to the incident ray $A B$, we must have
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Verified Answer
The correct answer is:
$\mu_4=\mu_1$
The correct option is $\mathbf{4} \mu_1=\mu_4$
Here, incident ray $A B$ is parallel to emergent ray $C D$
$\therefore \theta_1=\theta_4$
Since, incident ray $A B$ is parallel to emergent ray $C D$, so we can use
$\mu \sin \theta=$ constant
So, for first and last media,
$\begin{aligned} & \mu_1 \sin \theta_1=\mu_4 \sin \theta_4 \\ & \Rightarrow \mu_1=\mu_4 \quad\left(\because \theta_1=\theta_4\right)\end{aligned}$
Hence, option (4) is correct.
Here, incident ray $A B$ is parallel to emergent ray $C D$
$\therefore \theta_1=\theta_4$
Since, incident ray $A B$ is parallel to emergent ray $C D$, so we can use
$\mu \sin \theta=$ constant
So, for first and last media,
$\begin{aligned} & \mu_1 \sin \theta_1=\mu_4 \sin \theta_4 \\ & \Rightarrow \mu_1=\mu_4 \quad\left(\because \theta_1=\theta_4\right)\end{aligned}$
Hence, option (4) is correct.
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