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(A) According to Snell's law at the interface of the glass prism and the liquid at face $AC$:$\mu_{glass} \sin(i) = n \sin(r)$Given that the incident

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$27$

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(A) According to Snell's law at the interface of the glass prism and the liquid at face $AC$:$\mu_{glass} \sin(i) = n \sin(r)$Given that the incident angle $i = 60^{\circ}$ and the refracted light grazes along the face $AC$,the angle of refraction $r = 90^{\circ}$.Substituting the values:$1.5 	imes \sin(60^{\circ}) = n 	imes \sin(90^{\circ})$$1.5 	imes \frac{\sqrt{3}}{2} = n 	imes 1$$n = \frac{1.5 \sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}$We are given $n = \frac{\sqrt{x}}{4}$.Equating the two expressions for $n$:$\frac{\sqrt{x}}{4} = \frac{3 \sqrt{3}}{4}$$\sqrt{x} = 3 \sqrt{3} = \sqrt{9 	imes 3} = \sqrt{27}$Therefore,$x = 27$.

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