An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure.  The beaker height is 3h and its radius h.  When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod.  Then the refractive index of the liquid is

# An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure.  The beaker height is 3h and its radius h.  When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod.  Then the refractive index of the liquid is

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### Solution:

The line of sight of the observer remains constant, making an angle of 45° with the normal.

$\mathrm{sin\theta }=\frac{\mathrm{h}}{\sqrt{{\mathrm{h}}^{2}+{\left(2\mathrm{h}\right)}^{2}}}=\frac{1}{\sqrt{5}}$

$\mathrm{\mu }=\frac{\mathrm{sin}{45}^{\mathrm{o}}}{\mathrm{sin\theta }}=\frac{1/\sqrt{2}}{1/\sqrt{5}}=\sqrt{\left(\frac{5}{2}\right)}=1.58$

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