An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is N/100, where N is an integer, the value of N is _____.

Solution

Physics sir · 4 steps

The ray in air does not change, so find where it meets the liquid surface and how it bends there.

1. Empty jar. The jar is 2 × 15 = 30 cm wide. From the hole at 45 cm to the point at 15 cm on the far wall, the line drops 30 cm over 30 cm sideways. So the line of sight makes 45° with the vertical.
2. Ray in air. The observer looks along the same line, so the ray in air is still at i = 45°. From the hole at 45 cm down to the liquid surface at 30 cm it drops 15 cm, so it moves 15 cm sideways. It meets the surface 15 cm from the far wall.
3. Ray in liquid. From that point the ray must reach the bottom corner of the far wall: 15 cm sideways and 30 cm down. So tan r = 15/30 = 1/2, which gives sin r = 1/√5.

   μ = (sin i)/(sin r) = (1/√2)/(1/√5) = √(5/2) ≈ 1.58

4. So. μ = 1.58 = N/100, so N = 158.

The common slip: 200. Using the ratio of tangents, (tan 45°)/(tan r) = 2, gives 200. Snell's law needs the ratio of sines.

The idea to keep. When a liquid is added, keep the ray in air fixed. Split the path at the surface, get each angle from the horizontal and vertical distances, then apply Snell's law using sines.

Answer from the NTA final answer key. Solution by PrepWiser.