Question

Example 9.4: A crane flying 6 m above a still,clear water lake sees a fish underwater. Forthe crane, the fish appears to be 6 cm belowthe water surface. How much deep should thecrane immerse its beak to pick that fish?For the fish, how much above the water surfacedoes the crane appear? Refractive index ofwater = 4/3...​

Answer 1

Answer:

4.5 cm

Explanation:

$refractive \: index \: = \frac{real \: depth}{apprent \: depth}$

$let \: apparent \: depth = y$

$apparent \: depth \: for \: water = \frac{y}{3}$

$\frac{y}{3} = \frac{6{m}}{y}$

$y = \frac{6 \times 3}{4}$

$= \frac{18}{4}$

$= 4.5 \: m$

Answer 2

Answer:

R = 8 cm, A = 8 m

Explanation:

For crane, apparent depth of the fish is 6 cm and real depth is to be determined.

For fish, real depth (height, in this case) of the crane is 6 m and apparent depth (height) is to be determined.

n = $\frac{R}{A}$ = $\frac{Real\ depth} {Apparent\ depth}$

For crane, it is water with respect to air as real depth is in water and apparent depth is as seen from air

∴ n = $\frac{4}{3}$ = $\frac{R}{A}$ = $\frac{R}{6}$          ∴ R = 8 cm

For fish, it is air with respect to water as the real height is in air and seen from water.

∴ n = $\frac{3}{4}$ = $\frac{R}{A}$ = $\frac{6}{A}$          ∴ A = 8 m