Question

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 rifle 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)

Answer 1

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, $n=1.33$

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

$d'=h.tan\ \theta$

$d'=2\times tan\ 50$

$d'=2.3835\ m$

Now according to Snell's Law:

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

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

$\angle r=35.168^{\circ}$

Now the actual distance of the fish from the bottom surface at the normal:

$d=h.tan\ \theta$

$d=2\times tan\ 35.168$

$d=1.4092\ m$

Now distance between bullet hole and fish:

$\Delta d=d'-d$

$\Delta d=2.3835-1.4092$

$\Delta d=0.9743\ m$