Question

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

• A. movig 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

Answer 1

The correct option is A movig faster than its 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 $\frac{{\mu }_2}{v}=\frac{1}{u}$
or $v={\mu }_{2}u$
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
= refractive index × actual speed of bird.

Why this question? Tip: We should remember the formula of apparent depth for direct application