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course Phy 232
7/3 8:10PM
This experiment uses a cylindrical container and two lamps or other compact light sources. Fill a cylindrical container with water. The cylindrical section of a soft-drink bottle will suffice. The larger the bottle the better (e.g., a 2-liter bottle is preferable to a 20-oz bottle) but any size will suffice.
Position two lamps with bare bulbs (i.e., without the lampshades) about a foot apart and 10 feet or more from the container, with the container at the same height as the lamps. The line separating the two bulbs should be perpendicular to the line from one of the bulbs to the cylindrical container. The room should not be brightly lit by anything other than the two bulbs (e.g., don't do this in front of a picture window on a bright day).
The direction of the light from the bulbs changes as it passes into, then out of, the container in such a way that at a certain distance behind the container the light focuses. When the light focuses the images of the two bulbs will appear on a vertical screen behind the cylinder as distinct vertical lines. At the focal point the images will be sharpest and most distinct.
Using a book, a CD case or any flat container measure the distance behind the cylinder at which the sharpest image forms. Measure also the radius of the cylinder.
As explained in Index of Refraction using a Liquid and also in Class Notes #18, find the index of refraction of water.
Then using a ray-tracing analysis, as describe in Class Notes, answer the following:
1. If a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius then what is its angle of incidence on the cylinder?
Radius = 2 in.
Distance = 2 in * 1 / 4 = ½ in.
Angle of incidence = sin(angle of incidence) = ‘dX / r = ¼
Angle of incidence = 14.5 degrees
2. For the index of refraction you obtained, what therefore will be the angle of refraction for this ray?
Sin(angle of incidence) / sin(angle of refraction) = n_2 / n_1, where n is refraction of material.
We know n_1 is refraction of air = 1.0003 and n_2 is index of refraction of water = 1.33 so
Sin(14.5) = (1.33 / 1.0003) * sin( angle of refraction)
Sin(angle of refraction) = .188 so
Angle of refraction = 10.8 degrees
3. If this refracted ray continued far enough along a straight-line path then how far from the 'front' of the lens would it be when it crossed the central ray?
Since a ray of light parallel to the central ray strikes the cylinder at a distance equal to 1/4 of the cylinder's radius, this distance is equal to ½ in. therefore this should be equal to the distance from the ‘front’ of the lens where it is crossed with the central ray. This is calculated using triangles. The tan(angle of refraction) = ½ in / X
X = 2.62 in.
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The angle of refraction is between the radial line and the refracted ray, not between the refracted ray and the central axis. That angle is 14.5 degrees = 10.8 degrees = 3.8 degrees.
That will modify this result.
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4. How far from the 'front' of the lens did the sharpest image form?
2.5 in.
5. Should the answer to #3 be greater than, equal to or less than the answer to #4 and why?
These numbers should be equal, but are slightly different due to human error.
6. How far is the actual refracted ray from the central ray when it strikes the 'back' of the lens? What is its angle of incidence at that point? What therefore is its angle of refraction?
Since the lens is .25 in. We subtract the thickness from the distance we measured in number 4. So, 2.5 in - .25 in = 2.25 in. The angle of incidence would be the same as before equal to 14.5 degrees. The angle of refraction would still be the same, equal to 10.8 degrees.
7. At what angle with the central ray does the refracted ray therefore emerge from the 'back' of the lens?
This would be the same as the last part of number 6. The refraction angle is 10.8 degrees.
8. How far from the 'back' of the lens will the refracted ray therefore be when it crosses the central ray?
This would be the same as what we calculated in number 6, after subtracting the lens we get 2.25 in.
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Your trigonometry looks pretty good, but the refracted ray will actually reach the back of the lens before it crosses the central ray.
Check my note. The error can be easily corrected. Subsequent answers will also have to be adjusted.
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