Question

A fish is vertically below a flying bird moving vertically down towards water surface. The bird will appear to the fish to be ( assuming fish moves along the vertical joining fish and bird)

• A. moving faster than its real 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.

Answer 1

The correct option is A moving faster than its real speed and also away from the real distance.

$\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=\frac{{\mu }_2-{\mu }_1}{R}$
For a plane surface, $R=\infty$
$\therefore$ $\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v}=0$
or $\frac{{\mu }_2}{v}=\frac{{\mu }_1}{u}$ or $\frac{\mu }{v}=\frac{1}{u}$
or $v=\mu u$
Clearly, to the fish, the bird appears farther than its actual distance.
Again, $\frac{dv}{dt}=\mu \frac{du}{dt}$
or Apparent speed of bird
$=\text{refractive index}\times \text{actual speed of bird}$