A bird is flying $3 m$ above the surface of water of refractive index $4 / 3$ If the bird is diving vertically down with speed $6 m / s$ , his apparent velocity as seen by a stationary fish underwater (in $m / s$ ) is:

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Solution

Given
Distance of bird from surface $h = 3 m$
Refractive index of water $μ_{w a t e r} = 4 / 3$
Velocity of the bird $\frac{d h}{d t} = 6 m / s$

$\frac{μ_{a i r}}{h} = \frac{μ_{w a t e r}}{h_{i}}$
$\frac{1}{h} = \frac{μ_{w a t e r}}{h_{i}}$

Apparent height of bird fom surface
$h_{i}$ = $μ_{w a t e r} (h)$

Apparent velocity of bird as observed by fish = $\frac{d h_{i}}{d t} = μ_{w a t e r} \frac{d h}{d t} = \frac{4}{3} \times 6 = 8 m / s$

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A bird is flying 3 m above the surface of water of refractive index 4/3 If the bird is diving vertically down with speed 6 m/s, his apparent velocity as seen by a stationary fish underwater (in m/s) is:

Given Distance of bird from surface h=3 m Refractive index of water μwater=4/3 Velocity of the bird dhdt=6 m/s μairh=μwaterhi 1h=μwaterhi Apparent height of bird fom surface hi= μwater(h) Apparent velocity of bird as observed by fish = dhidt=μwaterdhdt=43×6=8 m/s

A bird is flying $3 m$ above the surface of water of refractive index $4 / 3$ If the bird is diving vertically down with speed $6 m / s$ , his apparent velocity as seen by a stationary fish underwater (in $m / s$ ) is: