A bird is flying at a height of $3.6 m$ above the surface of water and a fish is in the water at a depth of $1.2 m$ . The apparent height of bird to the fish is ( $μ_{w a t e r} = \frac{4}{3})$

3.9 m

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4.8 m

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5.2 m

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6.0 m

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Solution

The correct option is D $6.0 m$
Let apparent height of bird flying above water be $x$ ,
real depth / Apparent depth $= \frac{1}{μ} \Rightarrow \frac{3.6}{x} = \frac{3}{4}$
$\Rightarrow x = 4.8 m$
So,
Apparent height of bird to the fish is $= (4.8 + 1.2) m$
$= 6.0 m$

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A bird is flying at a height of 3.6m above the surface of water and a fish is in the water at a depth of 1.2m. The apparent height of bird to the fish is (μwater=43)

The correct option is D 6.0mLet apparent height of bird flying above water be x,real depth / Apparent depth =1μ⇒3.6x=34⇒x=4.8mSo,Apparent height of bird to the fish is =(4.8+1.2)m=6.0m

A bird is flying at a height of $3.6 m$ above the surface of water and a fish is in the water at a depth of $1.2 m$ . The apparent height of bird to the fish is ( $μ_{w a t e r} = \frac{4}{3})$

The correct option is D $6.0 m$
Let apparent height of bird flying above water be $x$ ,
real depth / Apparent depth $= \frac{1}{μ} \Rightarrow \frac{3.6}{x} = \frac{3}{4}$
$\Rightarrow x = 4.8 m$
So,
Apparent height of bird to the fish is $= (4.8 + 1.2) m$
$= 6.0 m$