Question #71966

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Question #71966

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Answer 1 · 2016-12-04T13:34:26

The fish is lying 1m below the surface of water and bird is flying 12 m above the surface of water. The fish is viewing the bird in air from denser medium water of refractive index $μ = \frac{4}{3}$ $mu=4/3$ .

If the bird is viewed nomally by the fish then we know the following relation.

$μ = \frac{Apparent depth in air}{Real depth in air}$ $mu="Apparent depth in air"/"Real depth in air"$

$\Rightarrow \frac{4}{3} = \frac{Apparent depth in air}{12 m}$ $=>4/3="Apparent depth in air"/(12m)$

$\Rightarrow Apparent depth in air = \frac{4}{3} \times 12 m = 16 m$ $=>"Apparent depth in air"=4/3xx12m=16m$

So the fish will see the bird at a distance $17 m$ $17m$ from it.

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Question #71966

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