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Index of Refraction

Date: 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      \

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   

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 

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 

                         \  \       |
                          \     \   |
                           \        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   
Associated Topics:
High School Physics/Chemistry

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