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Old May 19th 2009, 07:03 PM   #1
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More Snell's Law

During a training exercise, two British Navy officers used special equipment to escape from a submarine at a depth of 183m. Suppose that at a depth of 97m one of the officers shines a laser light toward the water's surface at an angle of incidence of 25 degrees. The light passes through the water and into the air, where it is detected by a rescue plane flying 44m above the water's surface. Calculate the horizontal distance between the rescue plane and the officers.

Okay. I got 109.976m. Is this correct?
I found it by using Snell's Law and constructing two right triangles and adding together their horizontal length.

Last edited by iEricKim; May 19th 2009 at 07:08 PM.
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Old May 19th 2009, 09:59 PM   #2
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Well I got 75.23 metres. To simplify things let us use the principle of reversibility of light. This means that if the laser were to be used by the pilot, the officers would see the
beam. You simply retrace the ray of light backwards.
Let the horizontal distance between the officers and the point where the light hits the
water surface be x and from there the horz distance to the plane be y. Let the angle of incidence of the reverse ray be i and that of the refracted ray be y. We have

1 x sin i = 1.333 sin 25 ( I used your value for water from your previous post) . This
gave me i = 34.2878. By complimentary angle thingy, the angle between the ray in the air and the ocean surface = 90 - 34.2878 = 55.712 Now, we have

44 / y = tan 55.712 or y = 30 metres.

Under water, we have x / 97 = tan 25 ( by alternate angle thingy) or

x = 97 tan 25 = 45.231 . Summing, x + y = 45.231 + 30 = 75.23 and hey presto
they are rescued!
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Old May 20th 2009, 01:47 AM   #3
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Originally Posted by physicsquest View Post
Well I got 75.23 metres. To simplify things let us use the principle of reversibility of light. This means that if the laser were to be used by the pilot, the officers would see the
beam. You simply retrace the ray of light backwards.
Let the horizontal distance between the officers and the point where the light hits the
water surface be x and from there the horz distance to the plane be y. Let the angle of incidence of the reverse ray be i and that of the refracted ray be y. We have

1 x sin i = 1.333 sin 25 ( I used your value for water from your previous post) . This
gave me i = 34.2878. By complimentary angle thingy, the angle between the ray in the air and the ocean surface = 90 - 34.2878 = 55.712 Now, we have

44 / y = tan 55.712 or y = 30 metres.

Under water, we have x / 97 = tan 25 ( by alternate angle thingy) or

x = 97 tan 25 = 45.231 . Summing, x + y = 45.231 + 30 = 75.23 and hey presto
they are rescued!
Lol. I forgot about the "complimentary angle thingy." xD
Thanks for the help.
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