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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 distance
μ1u+μ2v=μ2μ1R
For a plane surface, R=
μ1u+μ2v=0
or μ2vμ1u 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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