Snells Law (Changing view of an object through different materials)

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SUMMARY

This discussion centers on applying Snell's Law to determine the height of a shallow glass dish filled with water. The refractive indices used are n(water) = 1.333 and n(air) = 1.000. The solution involves using the tangent of angles rather than arcsin functions to relate the height of the dish to the dimensions of the triangles formed by the observer's line of sight. The final equation derived is 1.333 * sin(arctan(2/h)) = 1 * sin(arctan(4/h)), which simplifies the problem effectively.

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[SOLVED] Snells Law (Changing view of an object through different materials)

Homework Statement


A shallow glass dish is 4.0cm wide at the bottom, as shown in the attached figure. When at observer's eye is positioned as shown, the observer sees the edge of the bottom of the empty dish. When this dish is filled with water, the observer, with the eye positioned as before, sees the center of the bottom of the dish. Find the height of the dish.

Image:
img294.imageshack.us/img294/8576/prblmqp0.jpg

Homework Equations


Snell's Law:
n1 * sin(theta1) = n2 * sin(theta2)

n(water) = 1.333
n(air) = 1.000

The Attempt at a Solution


My attempt at the solution was futile. I can't seem to find a way to put the variable of height into Snell's law without being able to solve it. For example, I tried:

1.333(sin(arcsin(h/2)) = 1.000*(sin(arcsin(h/4))

But of course that would not work! How can I switch the variables up? Or am I trying to solve this the wrong way? Any help would be much appreciated.
 
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Hi rzermatt,

You had Snell's law as:

<br /> 1.333 \sin(\theta_i) = 1.000 \sin(\theta_r)<br />

and you saw that \theta_i was the top vertex of the triangle with bottom side of 2 cm.

Instead of inserting arc-functions, I think it might be better to follow this path: In terms of the sides of the triangle, what is \sin\theta_i? This will put the hypotenuse of each triangle in your equation, but go ahead and label them and put them in; then the trig functions will be gone.

Then, use the relationship between the hypotenuse of these triangles and their sides to get rid of the hypotenuse of each triangle. What do you get?


(By the way, the problem with the equation you gave is that it is the tangent of the angle that relates the sides h and 2, and sides h and 4, so your equation would need to be:

1.333 * sin(arctan(2/h) ) = 1 * sin(arctan(4/h)
 
That worked out great! I used the Pythagorean theorem to fix the relationships and then eliminate all the trig all together.

Thanks for the help!
 

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