A flying bird is stationary in the air vertically above a stationary fish in the water. If the fish appears to be at a distance $x_{1}$ to the bird and the bird appears to be at a distance $x_{2}$ to the fish then the correct choice is

x_{1} = x_{2}

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x_{1} > x_{2}

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x_{1} < x_{2}

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4 x_{2} < 3 x_{1}

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Solution

The correct option is C $x_{1} < x_{2}$
So, $x_{1}$ = apparent distance from bird to fish
$= y + \frac{x}{μ}$
and $x_{2}$ = apparent distance from fish to bird
$= μ y + x$
So, clearly $x_{1} < x_{2}$

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A flying bird is stationary in the air vertically above a stationary fish in the water. If the fish appears to be at a distance x1 to the bird and the bird appears to be at a distance x2 to the fish then the correct choice is

The correct option is C x1<x2So, x1 = apparent distance from bird to fish=y+xμand x2 = apparent distance from fish to bird=μy+xSo, clearly x1<x2

A flying bird is stationary in the air vertically above a stationary fish in the water. If the fish appears to be at a distance $x_{1}$ to the bird and the bird appears to be at a distance $x_{2}$ to the fish then the correct choice is

The correct option is C $x_{1} < x_{2}$
So, $x_{1}$ = apparent distance from bird to fish
$= y + \frac{x}{μ}$
and $x_{2}$ = apparent distance from fish to bird
$= μ y + x$
So, clearly $x_{1} < x_{2}$