Question

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:

• A. \( \sqrt{2} \)
• B. \( \frac{\sqrt{3}}{2} \)
• C. \( \sqrt{3} \)
• D. \( \frac{3}{2} \)

Hint:

To find the minimum refractive index, use the critical angle condition, where the angle of incidence equals the critical angle.

Answer 1

The Correct Option is A

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).