StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A coin rests at the bottom of an empty hemispherical bowl of radius R. An observer looking over the rim of the bowl can just barely not see the coin. When the bowl is completely filled with water (refractive index mu), the entire coin becomes just visible to the same observer in the same position. Find the diameter of the coin.

  1. 2R*(mu² - 1) / (mu² + 1)
  2. 2R*(mu - 1) / (mu + 1)
  3. R*(mu² - 1) / (mu² + 1)
  4. R*(mu - 1) / (mu + 1)

Correct answer: 2R*(mu² - 1) / (mu² + 1)

Solution

With an empty bowl, the line of sight from the rim just grazes the far edge of the coin. When water is added, refraction at the surface bends the ray so the coin's near edge also becomes visible. Applying Snell's law at the point where the ray exits the water surface at the rim, and using the hemispherical geometry (depth = R, radius = R), yields D = 2R*(mu²-1)/(mu²+1).

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →