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(D) The bird is at a height $3h$ from the bottom. The water level is at height $h$. The distance of the bird from the water surface is $d = 3h - h = 2

Vedclass · Accepted Answer

none of these

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(D) The bird is at a height $3h$ from the bottom. The water level is at height $h$. The distance of the bird from the water surface is $d = 3h - h = 2h$.First,the light from the bird refracts into the water. The apparent depth of the bird as seen from inside the water is $d' = \mu 	imes d = (4/3) 	imes 2h = 8h/3$.The total distance of this virtual object from the silvered bottom is $H = h + 8h/3 = 11h/3$.The mirror forms an image at a distance $H = 11h/3$ behind the mirror (below the bottom).This image acts as an object for the refraction at the water surface. The distance of this object from the surface is $H' = H + h = 11h/3 + h = 14h/3$.The final apparent position $y$ of the image as seen by the bird is $y = H' / \mu = (14h/3) / (4/3) = 14h/4 = 3.5h$.Since the water level $h$ is changing at $dh/dt = -1\, cm/s$,we differentiate $y$ with respect to time:$dy/dt = d/dt(3.5h) = 3.5 	imes (dh/dt) = 3.5 	imes (-1) = -3.5\, cm/s$.The negative sign indicates the image moves downwards at $3.5\, cm/s$.

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