A fish rising vertically upward with a uniform velocity of 8 ms⁻¹, observes that a bird is diving vertically downward towards the fish with the velocity of 12 ms⁻¹. If the refractive index of water is 4/3, then the actual velocity of the diving bird to pick the fish, will be __________ ms⁻¹.

Solution

Physics sir · 4 steps

From under water, the bird's height looks stretched by μ, so its speed looks stretched by μ too.

1. Apparent height. Let the bird be at a real height y above the water and the fish at depth x below it. The fish looks up from the denser medium, so the bird appears farther away, at a height μ y.
2. Apparent distance. The fish therefore sees the bird at a distance D = x + μ y. The fish's own depth x is real, so it is not scaled.

   dD/dt = dx/dt + μ dy/dt

3. Put in numbers. Both x and y are shrinking. The fish rises at 8 m s⁻¹ and the bird dives at v, so the apparent speed of approach is 8 + 4/3v = 12. This gives 4/3v = 4.
4. So. The actual speed of the bird is v = 4 × 3/4 = 3 m s⁻¹, so the answer is 3.

The common slip: 9. Ignoring the fish's own 8 m/s and writing 12 = 4/3v gives 9. The observed 12 m/s already includes the fish's motion.

The idea to keep. Write the apparent distance first, then differentiate it. Only the part in the other medium gets scaled: multiply by μ when you look from water into air, and divide by μ when you look from air into water.

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