A hemispherical vessel is completely filled with a liquid of refractive index \( \mu \). A small coin is kept at the lowest point \( O \) of the vessel as shown in the figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point \( E \) (at the level of the vessel) is:
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To find the minimum refractive index, use the critical angle condition, where the angle of incidence equals the critical angle.
To ensure visibility, the critical condition involves calculating the maximum angle of refraction, \(\theta_r\), formed when light exits the liquid at the surface.
The maximum angle for visibility is when light gets refracted horizontally, reaching the extreme of \(90^\circ\), thus the angle of incidence should not exceed the critical angle: \[ \sin(\theta_i) = \frac{1}{\mu} \Rightarrow \theta_i = \sin^{-1}\left(\frac{1}{\mu}\right) \]
The light path from \(O\) to \(E\) forms a right triangle where the hypotenuse equals the vessel's radius and the vertical leg aligns with the hemisphere. In this scenario, geometry dictates: \[ \sin(\theta_i) = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Equating the two relationships of \(\sin(\theta_i)\): \[ \frac{1}{\mu} = \frac{1}{\sqrt{2}} \]
Solve for \(\mu\): \[ \mu = \sqrt{2} \]
Thus, the minimum refractive index of the liquid for the coin to be visible from \(E\) is \(\sqrt{2}\)
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