Index of RefractionDate: 10/10/2001 at 14:01:45 From: Shelby Subject: Index of refraction Here is my question: At Norbert's backyard Halloween party, Dilbert, a guest, is considering jumping into Norbert's backyard swimming pool. Dilbert can't swim, but as he looks at the surface of the water, looking down at a 45-degree angle, the pool appears to be only 6 feet deep. "It's okay, I'm six feet, six inches tall!" Dilbert assures his host. "Hold it!" says Norbert. "It's deeper than you think!" How deep is the water? I know the index of refraction of air is 1, and 1.33 for water but I don't know what equation to plug them into to find the depth of the water. This is what I thought the correct equation is to solve for the other angle: n1 sin x1 = n2 sin x2 1.33 sin 45 degrees = sin x2 If this is correct, I came up with 70 degrees. But I don't know how to apply this in finding out the depth of the water! Am I way off course? Help! Thanks! Date: 10/10/2001 at 15:21:31 From: Doctor Douglas Subject: Re: Index of refraction Here's the diagram: D V A . | --------O----------E--- water surface |\ . | \ . | \ . | \ . | \ C | \ U \ \ \ \ B Dilbert looks into the water, and (assuming that the entire pool is filled with air) surmises that point C is 6 feet below point E. The dotted line (. . .) makes a 45-degree angle with the vertical line UV. In other words, the angle of incidence (angle DOV) is 45 degrees. But in fact D sees point B along the line formed by slashes (\). Dilbert can observe point B is directly below E, for example, by looking directly down on point E from point A. You are of course applying Snell's law: n1 sin(x1) = n2 sin(x2), but you have to make sure that the n1 goes with the x1. In this case the dotted line is along 45 degrees, and it is in the air, so x1 = 45 deg goes with n1 = 1.0 (index of refraction of air): 1.0 sin(45 deg) = 1.33 sin(x2) You can solve for x2, which is the angle that the ray from O to B makes with the vertical OU, i.e. angle UOB. Now you'll need to perform some trigonometry (knowing that the distance CE is 6 feet) to determine the actual depth (BE) of the pool. I hope that helps. Please write back if you need more explanation of what's going on. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 10/10/2001 at 16:25:02 From: Doctor Rick Subject: Re: Index of refraction Hi, Shelby. I see that Doctor Douglas answered your question. I'd like to add some explanation of why it works. It never seemed very obvious to me that an object on the bottom would appear to be directly above where it really is; but we can understand it by considering the psychophysics of depth perception. The big question is, how does Dilbert decide how deep the pool appears? The answer is, he uses his binocular vision. Consider an object just lying on the ground in front of you. The ray from that object to your left eye is in a vertical plane. The ray from the object to your right eye is in another vertical plane. It is the angle between these planes that your brain uses to determine distance: the smaller the distance, the more your eyes need to turn toward each other to look at the object. Knowing the distance between your eyes, and the angle between the planes, the brain can "calculate" the distance to the intersection of the two planes. This information, along with the angle of depression of either line of sight, is enough to compute the location of the object. We don't need to do the full calculation; we don't need to know the distance between the eyes, or the angle between the planes. All we have to note is that the intersection of the planes is a vertical line. Your brain finds this vertical line, then it finds the intersection of this line with the extension of the ray passing into the eye. That's where it thinks the object is. What happens when the object is at the bottom of the pool? Then the path followed by a light ray from the object to either eye is not a straight line - but the path is still in that same vertical plane. Therefore the true position of the object is on the same vertical line (the intersection of the planes) as the image - the place where Dilbert's brain tells him it is located. The figure now looks like this: EYE \ \ \ --------------------\-----------+----------------- \ \ | \ \ | \ O image \ | \ | \ | \ | \ | \ | \ | \| O object I have drawn the true (bent) path followed by a light ray, and the straight path that Dilbert's brain assumes the ray has followed. His brain thinks the object is at the place I labeled "image." - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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