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Question

# A fish is vertically below a flying bird moving vertically down towards water surface. The bird will appear to the fish to be

A
movig faster than its speed and also away from the real distance
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B
moving faster than its real speed and nearer than its real distance
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C
moving slower than its real speed and also nearer than its real distance
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D
moving slower than its real speed and away from the real distance
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Solution

## The correct option is A movig faster than its speed and also away from the real distanceμ1−u+μ2v=μ2−μ1R For a plane surface, R=∞ ∴ μ1−u+μ2v=0 or μ2v−μ1u or μ2v=1u or v=μ2u Clearly, to the fish, the bird appears farther than its actual distance. Again, dvdt=μdudt or Apparent speed of bird = refractive index × actual speed of bird. Why this question? Tip: We should remember the formula of apparent depth for direct application

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