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(C) Let the ray be incident normally on the left face. It enters the first prism (refractive index $\mu_1$) without deviation.At the interface between

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$\mu_1^2+\mu_3^2=2 \mu_2^2$

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(C) Let the ray be incident normally on the left face. It enters the first prism (refractive index $\mu_1$) without deviation.At the interface between prism $\mu_1$ and $\mu_2$,the angle of incidence is $i = 45^{\circ}$. By Snell's law: $\mu_1 \sin 45^{\circ} = \mu_2 \sin 	heta$,where $	heta$ is the angle of refraction.Thus,$\mu_1 \frac{1}{\sqrt{2}} = \mu_2 \sin 	heta \implies \sin 	heta = \frac{\mu_1}{\sqrt{2} \mu_2} \quad \dots(1)$At the interface between prism $\mu_2$ and $\mu_3$,the ray strikes at an angle of incidence $r_2 = \alpha - 	heta$. From the geometry,$\alpha = 90^{\circ}$. So,$r_2 = 90^{\circ} - 	heta$.The ray emerges normal to the right face,meaning the angle of refraction at the second interface is $0^{\circ}$. By Snell's law: $\mu_2 \sin(90^{\circ} - 	heta) = \mu_3 \sin 0^{\circ}$ is incorrect; rather,the ray emerges normal to the right face of the block,so the angle of incidence at the $\mu_2-\mu_3$ interface is $r_2$ and the angle of refraction is $i_2 = 45^{\circ}$.Applying Snell's law at the second interface: $\mu_2 \sin(90^{\circ} - 	heta) = \mu_3 \sin 45^{\circ} \implies \mu_2 \cos 	heta = \mu_3 \frac{1}{\sqrt{2}} \implies \cos 	heta = \frac{\mu_3}{\sqrt{2} \mu_2} \quad \dots(2)$Squaring and adding equations $(1)$ and $(2)$:$\sin^2 	heta + \cos^2 	heta = \left(\frac{\mu_1}{\sqrt{2} \mu_2}ight)^2 + \left(\frac{\mu_3}{\sqrt{2} \mu_2}ight)^2$$1 = \frac{\mu_1^2}{2 \mu_2^2} + \frac{\mu_3^2}{2 \mu_2^2} \implies 2 \mu_2^2 = \mu_1^2 + \mu_3^2$.

$A$ rectangular block is composed of three different glass prisms (with refractive indices $\mu_1, \mu_2$ and $\mu_3$) as shown in the figure below. $A$ ray of light incident normal to the left face emerges normal to the right face. Then,the refractive indices are related by

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