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.
The situation described is related to the critical angle and refraction in the liquid.
Step 1: For the person to see the coin from point \( E \), the light must undergo refraction at the surface of the liquid. The critical angle is given by: \[ \sin \theta_c = \frac{1}{\mu} \] where \( \theta_c \) is the critical angle.
Step 2: The refracted light must bend along the surface of the liquid. The angle of incidence at the surface should be equal to or greater than the critical angle.
Step 3: The coin is at the lowest point, and the person is at the level of the liquid. For this condition, the minimum refractive index \( \mu \) required is: \[ \mu = \sqrt{2} \]
Final Conclusion: The minimum refractive index of the liquid is \( \sqrt{2} \), which is Option (1).
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