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A swimmer is diving in a swimming pool v...

A swimmer is diving in a swimming pool vertically down with a velocity of `2 ms^(-I)`. What will be the velocity as seen by a stationary fish at the bottom of the pool right below the diver? Refractive index of water is 1.33

Text Solution

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In the figure, vertical distance 2 m is shown by AB. The height A from the surface of water is `h_0`, Suppose Its. apparent height is `h_i (h_i gt h_0)`
`therefore (h_i)/(h_0) = ("n(water)")/("n(air)")`
`therefore h_i= h_0 xx 1.33....... (1)`
Now the real height of B, `h_0 . = (h_0 + 2)m`
If it.s apparent height is `h_i`. then
`h._i/h._0 = ("n(water)")/("n(air)") = 1.33`
`therefore h_i = h_0 xx 1.33`
`therefore h_i = h_0 xx 1.33`
`= (h_0 + 2 ) xx 1.33`.....(2)

`rArr` From equations (1) and (2), the apparent distance seen by the fish
`h_1 - h_i = (h_0 +2) xx 1.33 - h_0 xx 1.33`
` = 2 xx 1.33 = 2.66`
`therefore` Apparent distance covered by fish means velocity.
`therefore` Apparent velocity seen by fish = 2.66 m/s
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