A bird is flying at a height of 2.4m above the surface of water and a fish is in the water at a depth of 1.2m. The apparent distance of the fish to the bird is :
( $μ_{w a t e r} = \frac{4}{3})$

3.0m

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3.3m

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1.2m

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0.9m

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Solution

The correct option is C 3.3m
Apparent depth of fish in the pond be $x$ ,
real depth / Apparent depth $= μ ∵$ the light moves from tenser to rarer medium
$\Rightarrow \frac{1.2}{x} = \frac{4}{3} \Rightarrow x = 0.9 m$
So,
The distance of fish to the bird $= (2.4 + 0.9) m$
$= 3.3 m$

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A bird is flying at a height of 2.4m above the surface of water and a fish is in the water at a depth of 1.2m. The apparent distance of the fish to the bird is :(μwater=43)

The correct option is C 3.3mApparent depth of fish in the pond be x,real depth / Apparent depth =μ∵ the light moves from tenser to rarer medium⇒1.2x=43⇒x=0.9mSo,The distance of fish to the bird =(2.4+0.9)m=3.3m

A bird is flying at a height of 2.4m above the surface of water and a fish is in the water at a depth of 1.2m. The apparent distance of the fish to the bird is :
( $μ_{w a t e r} = \frac{4}{3})$

The correct option is C 3.3m
Apparent depth of fish in the pond be $x$ ,
real depth / Apparent depth $= μ ∵$ the light moves from tenser to rarer medium
$\Rightarrow \frac{1.2}{x} = \frac{4}{3} \Rightarrow x = 0.9 m$
So,
The distance of fish to the bird $= (2.4 + 0.9) m$
$= 3.3 m$