A fish is swimming in a river at a depth of 20 cm from the surface of the water. Refractive Index of water is 4 3 . At what depth will an eagle flying above the water and exactly above the fish will see the fish? Options: A . &nbsp 10 cm B . &nbsp 20 cm C . &nbsp 26.66 cm D . &nbsp 15 cm

Answer: Option D : D Given: Refractive index of water, μ w a t e r = 4 3 Real depth of fish from surface of water = 20 c m In the given situation, Refractive Index of water = μ w a t e r = R e a l d e p t h A p p a r e n t d e p t h ⇒ A p p a r e n t d e p t h = R e a l d e p t h μ w a t e r ⇒ A p p a r e n t d e p t h = 20 4 3 Therefore, A p p a r e n t d e p t h = 20 × 3 4 = 15 cm. The eagle will perceive the fish to be at 15 c m depth from the water surface.

A fish is swimming in a river at a depth of 20 cm from the surface of the wate

Question

A fish is swimming in a river at a depth of 20 cm from the surface of the water. Refractive Index of water is $\frac{4}{3}$ . At what depth will an eagle flying above the water and exactly above the fish will see the fish?

Options:

A . 10 cm

B . 20 cm

C . 26.66 cm

D . 15 cm

Answer: Option D
:
D

Given:
Refractive index of water, $μ_{w a t e r}$ = $\frac{4}{3}$
Real depth of fish from surface of water $= 20 c m$
In the given situation,
Refractive Index of water $= μ_{w a t e r} = \frac{R e a l d e p t h}{A p p a r e n t d e p t h}$
$\Rightarrow A p p a r e n t d e p t h = \frac{R e a l d e p t h}{μ_{w a t e r}}$
$\Rightarrow A p p a r e n t d e p t h = \frac{20}{\frac{4}{3}}$
Therefore, $A p p a r e n t d e p t h = \frac{20 \times 3}{4}$ = 15 cm. The eagle will perceive the fish to be at $15 c m$ depth from the water surface.