
Given: Angle of incidence on the glass box is \(45^\circ\). The emergent ray along face AB implies the angle of refraction at the glass-liquid interface is \(90^\circ\). 1. Air-Glass Interface: \[ \frac{\sin 45^\circ}{\sin \theta} = \mu \] Substituting \(\sin 45^\circ = \frac{1}{\sqrt{2}}\): \[ \frac{1}{\sqrt{2}} = \mu \sin \theta \] 2. Glass-Liquid Interface: \[ \frac{\sin(90^\circ - \theta)}{\sin 90^\circ} = \frac{1}{\mu} \] Using \(\sin(90^\circ - \theta) = \cos \theta\) and \(\sin 90^\circ = 1\): \[ \cos \theta = \frac{1}{\mu} \] 3. Combining Equations: \[ \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\cos \theta} \] Simplifying gives: \[ \cos \theta = \sqrt{2} \sin \theta \] \[ \tan \theta = \frac{1}{\sqrt{2}} \] 4. Determining \(\sin \theta\): From triangle GEF: \[ \sin \theta = \frac{1}{\sqrt{3}} \] 5. Calculating Refractive Index (\(\mu\)): \[ \mu = \frac{1}{\sqrt{2} \sin \theta} = \frac{1}{\sqrt{2} \times \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}} \] The refractive index of the liquid is: \[ \boxed{\sqrt{\frac{3}{2}}} \]
