A bird B in air is diving vertically downwards over a water tank (made up of transparent glass) with speed 5 cm/s. Base of tank is silvered. A fish ‘F’ in the tank is rising vertically upwards along the same line with speed 2 cm/s. Water level in the tank is decreasing at the rate of 2 cm/s. $μ_{w a t e r} = \frac{4}{3},$ At that instant speed of image/images of fish as seen by the bird is/are (in cm/s)

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answer is C, D.

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Detailed Solution

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D) Direct image distance of fish wrt bird is
$X_{F B} = x_{1} + \frac{x_{2}}{^{μ}}$
$X_{F B} = \frac{d x_{1}}{d t} + \frac{1}{μ} \frac{d x_{2}}{^{d t}}$
= - (5 – 2)+ $\frac{3}{4}$ (-2 -2)
-3 + (- 3) = - 6 cm/s
C) mirror image distance of fish wrt bird is
$X_{F B} = x_{1} + \frac{h}{^{μ}} + \frac{h - x_{2}}{μ}$
$V_{F B} = \frac{d x_{1}}{d t} + \frac{2}{^{μ}} \frac{d h}{d t} - \frac{1}{μ} \frac{d x_{2}}{d t}$
$= - (5 - 2) + 2 \frac{3}{4} (- 2) - \frac{3}{4} (- 4)$
$= - 3 - 3 + 3 = - 3 c m / s$