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nlexa [21]
3 years ago
10

A catfish is spotted at the bottom of a creek. The depth of the water is 2 m. A hunter (with no knowledge of optics) fires a rif

le shot, directly aimed at the catfish. The bullet strikes the water at an angle of 40° to the water. Assume that the bullet is not deflected by the air/water interface, what is the distance from the bullet hole at the bottom of the creek to the catfish? Water: n=1.33. (Answer should be 0.97 m)

Physics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

\Delta d=0.9743\ m

Explanation:

Given:

  • depth of fish, h=2\ m
  • angle of incidence,\angle i=90-40=50^{\circ}
  • refractive index of water, n=1.33

<u>Apparent distance from the normal projection at the bottom of entrance at air-water surface to the fish:</u>

d'=h.tan\ \theta

d'=2\times tan\ 50

d'=2.3835\ m

<u>Now according to Snell's Law:</u>

n=\frac{sin\ i}{sin\ r}

1.33=\frac{sin\ 50}{sin\ r}

\angle r=35.168^{\circ}

<u>Now the actual distance of the fish from the bottom surface at the normal:</u>

d=h.tan\ \theta

d=2\times tan\ 35.168

d=1.4092\ m

<u>Now distance between bullet hole and fish</u>:

\Delta d=d'-d

\Delta d=2.3835-1.4092

\Delta d=0.9743\ m

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