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(B) Let the pinhole be at the origin. The light ray travels from the bottom of the rod to the pinhole and then to the observer's eye.From the geometry

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$\sqrt{\frac{5}{2}}$

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(B) Let the pinhole be at the origin. The light ray travels from the bottom of the rod to the pinhole and then to the observer's eye.From the geometry of the setup,the angle of incidence $	heta$ in the liquid is formed by the base $h$ and height $2h$ (as the rod is at the edge and the pinhole is at the center of the top surface).$\sin 	heta = \frac{h}{\sqrt{h^2 + (2h)^2}} = \frac{h}{\sqrt{5h^2}} = \frac{1}{\sqrt{5}}$.The angle of refraction in air is $45^\circ$ because the pinhole is at a horizontal distance $h$ from the rod and the vertical height above the liquid is $h$ (since total height is $3h$ and liquid level is $2h$).Using Snell's Law: $\mu \sin 	heta = 1 \cdot \sin 45^\circ$.$\mu \cdot \frac{1}{\sqrt{5}} = \frac{1}{\sqrt{2}}$.$\mu = \sqrt{\frac{5}{2}}$.

An observer can see the top of a thin rod of height $h$ through a pinhole. The height of the container is $3h$ and its radius is $2h$. When the container is filled with a liquid up to a height of $2h$,the observer can see the bottom end of the rod. Find the refractive index of the liquid.

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