A bird in air is at a height y from the surface of water s fish is at depth x below the surface of water the refractive index of water is mu the apperent distance of bird to the fish is
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A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 m/s observes a king fisher bird diving vertically towards water at the rate 9 m/s vertically above it. If the refractive index of water is 4/3, find the actual velocity of the dive of the bird.
Given,
Rate at which the fist is rising = 3 m/s
Rate at which the king fisher is diving = 9 m/s
Refractive index of water = 4/3

If at any instant, the fish is at a depth ‘x’ below water surface while, the bird is at a height 'y' above the surface, then the apparent height of the bird from the surface as seen by the fish will be given by
μ = Apparent heightReal heightμ = Apparent heightReal height
or, Apparent height = μy
So, the total apparent distance of the bird as seen by the fish in water will be h = x + μy
Now, differentiating w.r.t time, we get
dhdt=dxdt+μdydtdhdt=dxdt+μdydt
⇒⇒ 9 = 3+μ(dydt)9 = 3+μdydt
⇒⇒ dydt = 6(4/3) = 4.5 m/s.dydt = 6(4/3) = 4.5 m/s.
Thus, the kingfisher bird is diving with a velocity of 4.5 m/s.
Given,
Rate at which the fist is rising = 3 m/s
Rate at which the king fisher is diving = 9 m/s
Refractive index of water = 4/3

If at any instant, the fish is at a depth ‘x’ below water surface while, the bird is at a height 'y' above the surface, then the apparent height of the bird from the surface as seen by the fish will be given by
μ = Apparent heightReal heightμ = Apparent heightReal height
or, Apparent height = μy
So, the total apparent distance of the bird as seen by the fish in water will be h = x + μy
Now, differentiating w.r.t time, we get
dhdt=dxdt+μdydtdhdt=dxdt+μdydt
⇒⇒ 9 = 3+μ(dydt)9 = 3+μdydt
⇒⇒ dydt = 6(4/3) = 4.5 m/s.dydt = 6(4/3) = 4.5 m/s.
Thus, the kingfisher bird is diving with a velocity of 4.5 m/s.
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Answer:
apparent distance from bird to fish is U X+Y
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