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(B) The line of sight of the observer remains constant,making an angle of $45^{\circ}$ with the normal in the air.From the geometry of the setup,when

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$\sqrt{5/2}$

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(B) The line of sight of the observer remains constant,making an angle of $45^{\circ}$ with the normal in the air.From the geometry of the setup,when the liquid is filled to a height of $2h$,the light ray from the bottom of the rod travels to the liquid surface at a distance $h$ from the vertical axis of the beaker.The angle of incidence $	heta$ in the liquid is given by $	an 	heta = \frac{h}{2h} = 1/2$.Thus,$\sin 	heta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.Applying Snell's Law at the liquid-air interface: $\mu \sin 	heta = 1 \cdot \sin 45^{\circ}$.$\mu \left( \frac{1}{\sqrt{5}} ight) = \frac{1}{\sqrt{2}}$.Therefore,$\mu = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}}$.

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