Question

A bird flies down vertically towards a water surface. For a fish inside water, vertically below the bird, the bird will appear

• A. to be farther away than the actual distance.
• B. to be closer than the actual distance.
• C. to move faster than its actual speed
• D. to move slower than lts actual speed.

Answer 1

Answer: (A), (C)

The correct options are
A to be farther away than the actual distance.
C to move faster than its actual speed

If $\mu$ be the relative refractive index of the 2 materials,

then $\frac{RealHeight}{ApparentHeight}=\frac{1}{\mu }$

Thus, the apparent height of the bird will be more when viewed by the fish.

For refraction at plane surface$\frac{{\mu }_2}{v}=\frac{{\mu }_1}{\omega }$

Let x be the height of bird above water For light travelling from bird to fish
${\mu }_1=1;{\mu }_2=\mu$

$\omega =-x$

$\frac{\mu }{v}=-\frac{1}{x}$

$v=-\mu x$

$|v|=\mu x>x$

Speed of bird$=x$

Apparent speed of bird$=|v|$

$=\mu x$