A fish is swimming in a river at a depth of 20 cm from the surface of the water- Refractive Index of water is 4-3 - Then the eagle flying above the water will see the fish at

In the given situation, Refractive Index of water = Real depth/Apparent depth Refractive index of water, μ_water = 1/(3/4) = 4/3 Therefore, Apparent depth ...

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Physics

Apparent Depth

A fish is swi...

Question

A fish is swimming in a river at a depth of 20 cm from the surface of the water. Refractive Index of water is $\frac{4}{3}$ . Then the eagle flying above the water will see the fish at ___ .

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Solution

In the given situation,

Refractive Index of water
$= \frac{R e a l d e p t h}{A p p a r e n t d e p t h}$

Refractive index of water,
$μ_{w a t e r} = \frac{1}{(\frac{3}{4})} = \frac{4}{3}$

Therefore, Apparent depth
$= \frac{20 \times 3}{4} = 15 c m$

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A fish is swimming in a river at a depth of 20 cm from the surface of the water. Refractive Index of water is 43. Then the eagle flying above the water will see the fish at ___ .

In the given situation, Refractive Index of water =Real depthApparent depth Refractive index of water, μwater=1(34)=43 Therefore, Apparent depth =20×34=15 cm

A fish is swimming in a river at a depth of 20 cm from the surface of the water. Refractive Index of water is $\frac{4}{3}$ . Then the eagle flying above the water will see the fish at ___ .

In the given situation,

Refractive Index of water
$= \frac{R e a l d e p t h}{A p p a r e n t d e p t h}$

Refractive index of water,
$μ_{w a t e r} = \frac{1}{(\frac{3}{4})} = \frac{4}{3}$

Therefore, Apparent depth
$= \frac{20 \times 3}{4} = 15 c m$