A fish, F in the pond, is at a depth of 0.8 m from water surface and is moving vertically upwards with velocity 2 ms − 1 . At the same instant, a bird B is at a height of 6 m from water surface and is

Question

A fish, F in the pond, is at a depth of 0.8 m from water surface and is moving vertically upwards with velocity $2 ms^{- 1}$ . At the same instant, a bird B is at a height of 6 m from water surface and is moving downwards with velocity $3 ms^{- 1}$ . At this instant both are on the same vertical lines as shown in the figure. Which of the following statement(s) is(are) correct?

TestHub · Accepted Answer

If the observer is in water (fish looking at the bird): apparent distance $d = μx + y \Rightarrow v = μ \frac{d x}{d t} + \frac{d y}{d t}$

Height of B observed by F: apparent height above the surface = μ × 6 = 8 m, so from the fish (0.8 m deep) it is 8.8 m ≠ 8.00 m.

(A) is wrong.

If the observer is in air (bird looking at the fish): apparent depth = 0.8/μ = 0.6 m, so depth of F observed by B = 6 + 0.6 = 6.6 m.

(B) is correct.

Velocity of B observed by F: $v = μ \times 3 + 2 = 4 + 2 = 6 m / s \neq = 5 m / s$ (C) is wrong.

Velocity of F observed by B: $v = 3 + \frac{2}{μ} = 3 + 1.5 = 4.5 m / s$ (D) is correct.

Physics - Geometrical Optics Question with Solution | TestHub

Options:(select one or more)

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