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## Homework Statement

A foreign fishing submarine moves straight north through Canadian waters at a constant velocity of 12 m/s and a constant depth of 150m below the surface. A Canadian helicopter is in pursuit, flying in exactly the same direction at a constant velocity of 52 m/s and a constant altitude of 550m above the surface. The helicopter, lightly armed, shoots a narrow laser beam which enters the water and strikes the submarine. At one instant, the laser beam leaving the helicopter is directed at an angle of 10° below the horizontal to hit the target. At a later time, the beam has to be tilted down to 30° below the horizontal to maintain contact. Given that the index of refraction is 1.00 for air and 1.33 for water, find the time interval between the above events.

Possible Answers;

a) 33.3

b) 55.1

c) 68.9

d) 46.7

e) 78

## Homework Equations

n1xsin∅1 = n2xsin∅2

## The Attempt at a Solution

I found the two refractive angles;

1xsin10° = 1.33xsin∅1

∅1 = 7.5°

1xsin30° = 1.33xsin∅2

∅2 = 22.08°

After that, I'm stuck. I have no idea how to use that to find the time interval between each degree.

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