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

To ensure visibility, the critical condition involves calculating the maximum angle of refraction, \(\theta_r\), formed when light exits the liquid at the surface.

According to Snell's Law: \[ \mu \sin(\theta_i) = 1 \cdot \sin(90^\circ) = \mu \cdot \sin(\theta_i) \] Simplifying: \[ \sin(\theta_i) = \frac{1}{\mu} \]

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 \(\sqrt{2}\)