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Apparent velocity of fish floating in tank A tank is filled with water up to a height of 15cm and a fish is floating inside at a depth of 10cm. The tank is now being drained so that the surface moves down with a velocity of 1cm/s. An observer is viewing the fish from a height of 10cm. (a) Find the apparent velocity of the fish as seen by the observer. Find the apparent velocity of the observer as seen by the fish. |
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Answer» Solution :If the object moves with a velocity v inside the medium, its apparent velocity can be found by rate of change of apparent depth. Let us assume that `x_(m)=` depth of x with respect to surface. Then `(d_(app))/(n_("refracted"))=(d_("ACTUAL"))/(n_("incident"))` Here, the depth of image is changing because the depth of the object is changing with respect to surface. Differentiating both sides, we get `(V_(IS))/(n_(r ))=(V_(OS))/(n_(i))implies(V_(I)-V_(S))/(n_(r ))=(V_(O)-V_(S))/(n_(i))` where `V_(IS)` is the velocity of image with respect to the surface and `V_(OS)` is the object of image with respect to the surface. These refer to the velocities in the normal direction .In a direction parallel to the surface, the velocity of the image will be the same as the velocity of the object. Calculation : Assuming DOWNWARD direction to be positive, the velocity of the object (fish) is zero (fish is floating). Now `V_(S)=+1`cm/s. Therefore, `(V_(IS))/(n_(r ))=(V_(OS))/(n_(i))` `V_(IS)=(n_(r ))/(n_(i))V_(OS)=(1)/((4)/(3))xx(V_(O)-V_(S))-(3)/(4)m//s` `V_(IS)=V_(I)-V_(S)` `-(3)/(4)+1=V_(I)impliesV_(I)=(1)/(4)m//s` (34-5) Again `(V_(IS))/((4)/(3))=(V_(OS))/(1)=V_(S)xx(4)/(3)=(-4)/(3)m//s` `V_(I)-V_(S)=(4)/(3)` `V_(I)=1-(-4)/(3)=(-1)/(3)cm//s` Learn : The fish seems to be at a lesser depth than it actually is. So when depth of water DECREASES, the image of fish will approach its actual position. So, `V_(I)` should be downward in EQ. 34-5. A similar argument can be applied to Eq. 34-6 |
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