SATHEE: Optics Question 455

Optics Question 455

Question: A fish is vertically below a flying bird moving vertically= down towards water surface. The bird will appear to the fish to be

Options:

A) Moving faster than its speed and also away from the real distance

B) Moving faster than its real speed and nearer than its real distance

C) Moving slower than its real speed and also nearer than its real distance

D) Moving slower than its real speed and away from the real distance

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $\frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = \frac{μ_{1} - μ_{2}}{R}$ For a plane surface, $R = \infty$

$∴ \frac{μ_{1}}{- u} + \frac{μ_{2}}{v} = 0$

or $\frac{μ_{2}}{v} = \frac{μ_{1}}{u}$

or $\frac{μ}{v} = \frac{1}{u}$

Or $v = μ u$

Clearly, to the fish, the bird appears farther than its actual distance. Again, $\frac{d v}{d t} = μ \frac{d u}{d t}$

Or. Apparent speed of bird = refractive index $\times$ Actual speed of bird

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Optics Question 455

Question: A fish is vertically below a flying bird moving vertically= down towards water surface. The bird will appear to the fish to be

Options:

Answer:

Solution:

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