A fish inside a pond appears to be at a depth of 1.5m when viewed vertically from above. The refractive index…

Analyze the question

A fish in water is viewed from above. Its apparent depth is 1.5 m, and the refractive index of water is 1.3. We need the apparent displacement, which is the difference between the real depth and the apparent depth.

Concept

For an object viewed vertically from above:

refractive index n = real depth / apparent depth

Apparent displacement = real depth − apparent depth

Solution

1. Real depth = n × apparent depth = 1.3 × 1.5 = 1.95 m
2. Apparent displacement = real depth − apparent depth = 1.95 − 1.5 = 0.45 m

The fish seems to be lifted 0.45 m closer to the surface than it really is.

Answer

The apparent displacement is 0.45 m, which is option D.

Tip

Be careful when the question says "appears to be at a depth of..." because that value is the apparent depth, not the real depth. Many students wrongly pick 1.95 m (option B), which is the real depth, not the displacement. Always check whether the question wants depth or displacement. A quick check: the displacement should be a fraction of the real depth, equal to real depth × (1 − 1/n).