A spherical aquarium of radius 10 cm filled with water of refractive index 4/3 is kept as shown. One half of the aquarium is silvered and the other half is smooth. The thickness of the aquarium is negligible. A fish is moving towards left with a velocity 6 m/s. An observer detects two images of the fish. Find the relative velocity (in m/s) between the two images of the fish as seen by the observer.

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answer is 16.

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Detailed Solution

For reaction only
$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R} .... (i) v = - 10 c m$
Differentiate equation (i) w.r.t. time
$- \frac{μ_{2}}{v^{2}} \frac{d v}{d t} + \frac{μ_{1}}{d t} = 0$
$\frac{d v}{d t} = \frac{μ_{1}}{μ^{2}} \frac{v^{2}}{μ^{2}} + \frac{d u}{d t} = \frac{4}{1 \times 3} \times 1 \times 6 = 8 m / s$ (towards left)
The velocity of image formed after refraction is 8 m/s (towards left)
The velocity of image formed after reflection and then refraction =8 m/s (towards right).
The relative velocity between two images formed $= 16 m / s$