A stick of height $$h$$ is protruding out of a liquid contained in a large vessel with half of it inside the liquid. Rays of light from after noon sun are incident on the stick at $$45^\circ$$ as shown by the dotted lines. Find the length of the shadow on the bottom of the vessel. Would the shadow length increase or decrease as the sun goes down further?
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The ray grazing the top of the stick meets the liquid surface at a horizontal distance $$\frac{h}{2}\tan 45^\circ = \frac{h}{2}$$ from the stick. There it refracts with $$\sin 45^\circ = n\sin r$$, so $$\tan r = \frac{1}{\sqrt{2n^2 - 1}}$$, and in the depth $$\frac{h}{2}$$ it moves a further $$\frac{h}{2}\tan r$$ sideways. The shadow on the bottom is therefore $$\frac{h}{2}\left(1 + \frac{1}{\sqrt{2n^2 - 1}}\right)$$; taking the liquid as water with $$n = \frac{4}{3}$$, $$\tan r = \frac{3}{\sqrt{23}}$$ and the shadow is about $$0.81h$$. As the sun goes down the angle of incidence grows, so both $$\tan i$$ and $$\tan r$$ increase and the shadow becomes longer.