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

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moving faster than its actual speed and also away from the real distance

moving faster than its real speed and nearer than its real distance.

moving slower than its real speed and away from the real distance.

moving slower than its real speed and also nearer than its real distance.

answer is A.

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Detailed Solution

$\frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$For a plane surface, R = \infty$

$∴ \frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = 0$

or $\frac{μ_{2}}{v} = \frac{μ_{1}}{u} or \frac{μ}{v} = \frac{1}{u}$

or $v = μ_{2} u$

Clearly, to the fish, the bird appears farther than its actual distance.

$Again, \frac{d v}{d t} = μ \frac{d u}{d t}$

or Apparent speed of bird = Refractive index x Actual speed of bird.

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